Question 1: Integrate x \sin x
Solution:
I = \int x \sin x dx
Take u = x v = \sin x
Differentiate u: \dfrac{du}{dx} = 1
Integrate v: \int v \cdot dx = \int \sin x dx = -\cos x
By parts formula: \int u \cdot v \cdot dx = u \int v dx - \int \frac{du}{dx} \int v \cdot dx \cdot dx
Substitute values: \int x \sin x dx = x \cdot (-\cos x) - \int 1 \cdot (-\cos x), dx
I = -x \cos x + \int \cos x dx
I = -x \cos x + \sin x + C
Final Answer: \boxed{\int x \sin x dx = -x \cos x + \sin x + C}
Question 2: Integrate x \sin 3x
Solution:
I = x \sin 3x
Take u = x v = \sin 3x
Differentiate u: \dfrac{du}{dx} = 1
Integrate v: \int v \cdot dx = \int \sin 3x, dx = -\frac{1}{3}\cos 3x
By parts formula: \int u \cdot v \cdot dx = u \int v dx - \int \frac{du}{dx} \int v \cdot dx \cdot dx
Substitute values:
\int x \sin 3x, dx = x \cdot \left(-\frac{1}{3} \cos 3x\right) - \int 1 \cdot \left(-\frac{1}{3} \cos 3x\right), dx
I = -\frac{x}{3} \cos 3x + \frac{1}{3} \int \cos 3x, dx
I = -\frac{x}{3} \cos 3x + \frac{1}{3} \cdot \frac{1}{3} \sin 3x
I = -\frac{x}{3} \cos 3x + \frac{1}{9} \sin 3x + C
Final Answer: \boxed{ \int x \sin 3x, dx = -\dfrac{x}{3} \cos 3x + \dfrac{1}{9} \sin 3x + C}
Question 3: Integrate x^2 e^x
Solution:
I = \int x^2 e^x \cdot dx
Take u = x^2 v = e^x
Differentiate u: \dfrac{du}{dx} = 2x
Integrate v: \int v \cdot dx = \int e^x, dx = e^x
By parts formula: \int u \cdot v \cdot dx = u \int v dx - \int \frac{du}{dx} \int v \cdot dx \cdot dx
Substitute values: I = x^2 e^x - \int 2x e^x, dx
Now, solve \int 2x e^x \cdot dx using by parts again:
Take u_1 = 2x v_1 = e^x
Differentiate u_1: \dfrac{du_1}{dx} = 2
Integrate v_1: \int v \cdot dx = \int e^x \cdot dx = e^x
Applying by parts: \int 2x e^x \cdot dx = 2x e^x - \int 2 e^x \cdot dx = 2x e^x - 2 e^x
Substitute back:
I = x^2 e^x - [2x e^x - 2 e^x]
I = x^2 e^x - 2x e^x + 2 e^x
I = e^x (x^2 - 2x + 2) + C
Final Answer: \boxed{ \int x^2 e^x \cdot dx = e^x (x^2 - 2x + 2) + C}
Question 4: Integrate x \cdot \log x
Solution:
I = x \cdot \log x
Take u = \log x v = x
Differentiate u: \dfrac{du}{dx} = \dfrac{1}{x}
Integrate v: \int x \cdot dx = \dfrac{x^2}{2}
By parts formula: \int u \cdot v \ dx = u \int v \ dx - \int \dfrac{du}{dx} \int v \ dx \ dx
Substitute values: I = \log x \cdot \dfrac{x^2}{2} - \int \dfrac{1}{x} \cdot \dfrac{x^2}{2} \ dx
I = \dfrac{x^2}{2} \log x - \dfrac{1}{2} \int x \ dx
I = \dfrac{x^2}{2} \log x - \dfrac{1}{2} \cdot \dfrac{x^2}{2} + C
Final Answer: \boxed{\int x \cdot \log x \ dx = \dfrac{x^2}{2} \log x - \dfrac{x^2}{4} + C}
Question 5: Integrate x \cdot \log 2x
Solution:
I = \int x \cdot \log 2x \ dx
We know \log 2x = \log 2 + \log x.
So, x \cdot \log 2x = x \cdot \log 2 + x \cdot \log x
Integrate term by term:
I = \int x \cdot \log 2 \ dx + \int x \cdot \log x \ dx
\log 2 is a constant, so
\log 2 \int x \ dx = \log 2 \cdot \dfrac{x^2}{2}
From Question 4, \int x \cdot \log x, dx = \dfrac{x^2}{2} \log x - \dfrac{x^2}{4}
So, \int x \cdot \log 2x, dx = \log 2 \cdot \dfrac{x^2}{2} + \dfrac{x^2}{2} \log x - \dfrac{x^2}{4} + C
This can also be written as:
I = \dfrac{x^2}{2} (\log 2 + \log x) - \dfrac{x^2}{4} + C
I = \dfrac{x^2}{2} \log (2x) - \dfrac{x^2}{4} + C
Final Answer: \boxed { \int x \cdot \log 2x \ dx = \dfrac{x^2}{2} \log (2x) - \dfrac{x^2}{4} + C }
Question 6: Integrate x^2 \cdot \log x
Solution:
I = \int x^2 \cdot \log x \ dx
Take u = \log x v = x^2
Differentiate u: \dfrac{du}{dx} = \dfrac{1}{x}
Integrate v: \int x^2 \cdot dx = \dfrac{x^3}{3}
By parts formula: I = u \int v \ dx - \int \dfrac{du}{dx} \int v \ dx \ dx
Substitute values: I = \log x \cdot \dfrac{x^3}{3} - \int \dfrac{1}{x} \cdot \dfrac{x^3}{3} \ dx
I = \dfrac{x^3}{3} \log x - \dfrac{1}{3} \int x^2 \ dx
I = \dfrac{x^3}{3} \log x - \dfrac{1}{3} \cdot \dfrac{x^3}{3} + C
I = \dfrac{x^3}{3} \log x - \dfrac{x^3}{9} + C
Final Answer: \boxed{\int x^2 \cdot \log x \ dx = \dfrac{x^3}{3} \log x - \dfrac{x^3}{9} + C}
Question 7: Integrate x \cdot \sin^{-1} x
Solution:
I = \int x \cdot \sin^{-1} x \cdot dx
Take u = \sin^{-1} x v = x
Differentiate u: \dfrac{du}{dx} = \dfrac{1}{\sqrt{1-x^2}}
Integrate v: \int v.dx = \int x \cdot dx = \dfrac{x^2}{2}
By parts formula: \int u \cdot v \ dx = u \int v \ dx - \int \dfrac{du}{dx} \int v \ dx \ dx
Substitute values:
I = \int x \cdot \sin^{-1} x \ dx
I = \sin^{-1} x \cdot \dfrac{x^2}{2} - \int \dfrac{1}{\sqrt{1-x^2}} \cdot \dfrac{x^2}{2} \ dx
Take \dfrac{1}{2} outside:
I = \dfrac{x^2}{2} \sin^{-1} x - \dfrac{1}{2} \int \dfrac{x^2}{\sqrt{1-x^2}} \ dx
Now, x^2 = (1 - (1 - x^2)) = (1 - (1 - x^2)) = 1 - (1 - x^2)
So, \dfrac{x^2}{\sqrt{1-x^2}} = \dfrac{1}{\sqrt{1-x^2}} - \dfrac{1 - x^2}{\sqrt{1-x^2}}
So, \int \dfrac{x^2}{\sqrt{1-x^2}} \ dx = \int \dfrac{1}{\sqrt{1-x^2}} \ dx - \int \dfrac{1-x^2}{\sqrt{1-x^2}} \ dx
So, \int \dfrac{x^2}{\sqrt{1-x^2}} \ dx = \int \dfrac{1}{\sqrt{1-x^2}} \ dx - \int \sqrt{1-x^2} \ dx
Substituting,
I = \dfrac{x^2}{2} \sin^{-1} x - \int \dfrac{1}{\sqrt{1-x^2}} \ dx - \int \sqrt{1-x^2} \ dx
Using formulas,
\int \dfrac{1}{\sqrt{1-x^2}}, dx = \sin^{-1} x
\int \sqrt{1-x^2}, dx = \dfrac{x}{2} \sqrt{1-x^2} + \dfrac{1}{2} \sin^{-1} x
Putting everything together:
I = \int x \cdot \sin^{-1} x, dx = \dfrac{x^2}{2} \sin^{-1} x - \dfrac{1}{2} \left( \sin^{-1} x - \left( \dfrac{x}{2} \sqrt{1- x^2} + \dfrac{1}{2} \sin^{-1} x \right) \right) + C
Simplify:
I = \dfrac{x^2}{2} \sin^{-1} x - \dfrac{1}{2} \sin^{-1} x + \dfrac{x}{4} \sqrt{1-x^2} + \dfrac{1}{4} \sin^{-1} x + C
I = (\dfrac{x^2}{2} - \dfrac{1}{2} + \dfrac{1}{4}) \sin^{-1} x + \dfrac{x}{4} \sqrt{1-x^2} + C
I = \dfrac{2x^2 - 2 + 1}{4} \sin^{-1} x + \dfrac{x}{4} \sqrt{1-x^2} + C
I = \dfrac{2x^2 - 1}{4} \sin^{-1} x + \dfrac{x}{4} \sqrt{1-x^2} + C
Final Answer: \boxed{ \dfrac{2x^2 - 1}{4} \sin^{-1} x + \dfrac{x}{4} \sqrt{1-x^2} + C }
Question 8: Integrate x \cdot \tan^{-1} x
Solution:
I = \int x \cdot \tan^{-1} x \ dx
Take u = \tan^{-1} x v = x
Differentiate u: \dfrac{du}{dx} = \dfrac{1}{1 + x^2}
Integrate v: \int x \cdot dx = \dfrac{x^2}{2}
By parts formula: \int u \cdot v , dx = u \int v, dx - \int \dfrac{du}{dx} \int v, dx , dx
Substitute values: I = \tan^{-1} x \cdot \dfrac{x^2}{2} - \int \dfrac{1}{1 + x^2} \cdot \dfrac{x^2}{2} \ dx
Take \dfrac{1}{2} outside:
I = \dfrac{x^2}{2} \tan^{-1} x - \dfrac{1}{2} \int \dfrac{x^2}{1 + x^2} \ dx
Now, \dfrac{x^2}{1 + x^2} = 1 - \dfrac{1}{1 + x^2}
So, \int \dfrac{x^2}{1 + x^2} \ dx = \int 1 \ dx - \int \dfrac{1}{1 + x^2} \ dx = x - \tan^{-1} x
Putting it all together:
I = \dfrac{x^2}{2} \tan^{-1} x - \dfrac{1}{2} (x - \tan^{-1} x) + C
I = \dfrac{x^2}{2} \tan^{-1} x - \dfrac{x}{2} + \dfrac{1}{2} \tan^{-1} x + C
Final Answer: \boxed{\int x \cdot \tan^{-1} x \ dx = \dfrac{x^2}{2} \tan^{-1} x - \dfrac{x}{2} + \dfrac{1}{2} \tan^{-1} x + C }
Question 9: Integrate x \cdot \cos^{-1} x
Solution:
I = \int x \cdot \cos^{-1} x \ dx
Take u = \cos^{-1} x v = x
Differentiate u: \dfrac{du}{dx} = -\dfrac{1}{\sqrt{1-x^2}}
Integrate v: \int x \cdot dx = \dfrac{x^2}{2}
By parts formula: latex]\int u \cdot v / dx = u \int v, dx – \int \dfrac{du}{dx} \int v / dx / dx[/latex]
Substitute values: I = \cos^{-1} x \cdot \dfrac{x^2}{2} - \int \left( -\dfrac{1}{\sqrt{1-x^2}} \cdot \dfrac{x^2}{2} \right) \ dx
I= \dfrac{x^2}{2} \cos^{-1} x + \dfrac{1}{2} \int \dfrac{x^2}{\sqrt{1-x^2}} \ dx
Similar to question 7,
\int \dfrac{x^2}{\sqrt{1-x^2}} dx = \int \dfrac{1}{\sqrt{1-x^2}} dx - \int \sqrt{1-x^2} dx
So,
I = \dfrac{x^2}{2} \cos^{-1} x + \dfrac{1}{2} \left( \int \dfrac{1}{\sqrt{1-x^2}} dx - \int \sqrt{1-x^2} dx \right)
We know: \int \dfrac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x
\int \sqrt{1-x^2} dx = \dfrac{x}{2} \sqrt{1-x^2} + \dfrac{1}{2} \sin^{-1} x
So,
I = \dfrac{x^2}{2} \cos^{-1} x + \dfrac{1}{2} \left( \sin^{-1} x - \left( \dfrac{x}{2} \sqrt{1-x^2} + \dfrac{1}{2} \sin^{-1} x \right) \right) + C
I = \dfrac{x^2}{2} \cos^{-1} x + \dfrac{1}{2} \sin^{-1} x - \dfrac{x}{4} \sqrt{1-x^2} - \dfrac{1}{4} \sin^{-1} x + C
I = \dfrac{x^2}{2} \cos^{-1} x + \dfrac{1}{4} \sin^{-1} x - \dfrac{x}{4} \sqrt{1-x^2} + C
Final Answer: \boxed{ \int x \cdot \cos^{-1} x \ dx = \dfrac{x^2}{2} \cos^{-1} x + \dfrac{1}{4} \sin^{-1} x - \dfrac{x}{4} \sqrt{1-x^2} + C }
Question 10: Integrate (\sin^{-1} x)^2
Solution:
I = \int (\sin^{-1} x)^2 \ dx
Take u = (\sin^{-1} x)^2 v = 1
Differentiate u: \dfrac{du}{dx} = 2 \sin^{-1} x \cdot \dfrac{1}{\sqrt{1-x^2}}
Integrate v: \int 1 \cdot dx = x
By parts formula: \int u \cdot v \ dx = u \int v \ dx - \int \dfrac{du}{dx} \int v \ dx \ dx
Substitute values:
I = (\sin^{-1} x)^2 \cdot x - \int \left[2 \sin^{-1} x \cdot \dfrac{1}{\sqrt{1-x^2}}\right] \cdot x \ dx
I = x (\sin^{-1} x)^2 - 2 \int \dfrac{x \cdot \sin^{-1} x}{\sqrt{1-x^2}} \ dx
Integrating: \int \dfrac{x \cdot \sin^{-1} x}{\sqrt{1-x^2}} \ dx
Let’s put t = \sin^{-1} x \implies x = \sin t, , dx = \cos t \cdot dt
So, x = \sin t, \qquad \sin^{-1} x = t, \qquad \sqrt{1-x^2} = \cos t
So, \dfrac{x \cdot \sin^{-1} x}{\sqrt{1-x^2}} \cdot dx = \dfrac{\sin t \cdot t}{\cos t} \cdot \cos t , dt = t \sin t , dt
So, \int \dfrac{x \cdot \sin^{-1} x}{\sqrt{1-x^2}} , dx = \int t \sin t , dt
Now,
Integrate t \sin t , dt by parts:
Take u = t v = \sin t
Differentiate u: \dfrac{du}{dx} =1
Integrate v: \int \sin t \ dx = -\cos t
By parts:
\int t \sin t \ dt = t \cdot (-\cos t) - \int 1 \cdot (-\cos t) \ dt = -t \cos t + \int \cos t \ dt = -t \cos t + \sin t
Now, substitute back t = \sin^{-1} x, \cos t = \sqrt{1-x^2}, \sin t = x:
So, \int \dfrac{x \cdot \sin^{-1} x}{\sqrt{1-x^2}} \ dx = -\sin^{-1} x \cdot \sqrt{1-x^2} + x
Now put back into the original integral:
I = x (\sin^{-1} x)^2 - 2 \left(-\sin^{-1} x \cdot \sqrt{1-x^2} + x\right) + C
I = x (\sin^{-1} x)^2 + 2 \sin^{-1} x \cdot \sqrt{1-x^2} - 2x + C
Final Answer: \boxed{\int (\sin^{-1} x)^2 \ dx = x (\sin^{-1} x)^2 + 2 \sin^{-1} x \cdot \sqrt{1-x^2} - 2x + C}
Question 11: Integrate \dfrac{x \cdot \cos^{-1} x}{\sqrt{1-x^2}}
Solution:
Let I = \int \dfrac{x \cdot \cos^{-1} x}{\sqrt{1-x^2}} \cdot dx
Let t = \cos^{-1} x,
so x = \cos t,
differentiating dx = -\sin t , dt,
also \sqrt{1-x^2} = \sin t
Substitute:
I = \int \dfrac{\cos t \cdot t}{\sin t} \cdot (-\sin t \ dt)
I= -\int t \cos t \ dt
Now, integrate \int t \cos t \ dt by parts:
Take
u = t v = \cos t
Differentiate u: \dfrac{du}{dt} = 1
Integrate v: \int v \ dt = \sin t
By parts: \int t \cos t \ dt = t \sin t - \int 1 \cdot \sin t \ dt = t \sin t + \cos t
So, I = -\left(t \sin t + \cos t\right) + C
Now, substitute back t = \cos^{-1} x, \sin t = \sqrt{1-x^2}, \cos t = x:
So, I = -\left(\cos^{-1} x \cdot \sqrt{1-x^2} + x\right) + C
Final Answer: \boxed{ \int \dfrac{x \cdot \cos^{-1} x}{\sqrt{1-x^2}} dx = -\cos^{-1} x \cdot \sqrt{1-x^2} - x + C }
Question 12: Integrate x \cdot \sec^2 x
Solution:
Let I = \int x \cdot \sec^2 x \ dx
Take u = x v = \sec^2 x
Differentiate u: \dfrac{du}{dx} = 1
Integrate v: \int \sec^2 x \ dx = \tan x
By parts formula: \int u \cdot v \ dx = u \int v \ dx - \int \dfrac{du}{dx} \int v \ dx \ dx
Substitute values:
I = x \cdot \tan x - \int 1 \cdot \tan x \ dx
I = x \cdot \tan x - \int \tan x , dx
I = x \cdot \tan x - \int \dfrac{\sin x}{\cos x} \ dx
I = x \cdot \tan x + \log |\cos x| + C
Final Answer: \boxed{ \int x \cdot \sec^2 x \ dx = x \cdot \tan x + \log |\cos x| + C}
Question 13: Integrate \tan^{-1} x
Solution:
Let I = \int \tan^{-1} x \ dx
Take u = \tan^{-1} x v = 1
Differentiate u: \dfrac{du}{dx} = \dfrac{1}{1 + x^2}
Integrate v: \int v \ dx = x
By parts formula: \int u \cdot v \ dx = u \int v \ dx - \int \dfrac{du}{dx} \int v \ dx \ dx
Substitute values:
I = \tan^{-1} x \cdot x - \int \dfrac{1}{1 + x^2} \cdot x \ dx
I = x \cdot \tan^{-1} x - \int \dfrac{x}{1 + x^2} \ dx
Integrating I = \int \dfrac{x}{1 + x^2} \ dx
Let t = 1 + x^2 \implies dt = 2x \cdot dx
So, \int \dfrac{x}{1 + x^2} \ dx = \dfrac{1}{2} \int \dfrac{1}{t} dt = \dfrac{1}{2} \log |1 + x^2|
So, the integral is:
I = x \cdot \tan^{-1} x - \dfrac{1}{2} \log (1 + x^2) + C
Final Answer: \boxed{ \int \tan^{-1} x \ dx = x \cdot \tan^{-1} x - \dfrac{1}{2} \log (1 + x^2) + C }
Question 14: Integrate x \cdot (\log x)^2
Solution:
Let I = \int x \cdot (\log x)^2 \ dx
Take u = (\log x)^2 v = x
Differentiate u: \dfrac{du}{dx} = 2 \log x \cdot \dfrac{1}{x} = \dfrac{2 \log x}{x}
Integrate v: \int x \cdot dx = \dfrac{x^2}{2}
By parts formula: \int u \cdot v \ dx = u \int v \ dx - \int \dfrac{du}{dx} \int v \ dx \ dx
Substitute values:
I = (\log x)^2 \cdot \dfrac{x^2}{2} - \int \dfrac{2 \log x}{x} \cdot \dfrac{x^2}{2} dx
I = \dfrac{x^2}{2} (\log x)^2 - \int \log x \cdot x \ dx
Now, from Question 4:
\int x \cdot \log x / dx = \dfrac{x^2}{2} \log x - \dfrac{x^2}{4}
So, I = \dfrac{x^2}{2} (\log x)^2 - \left( \dfrac{x^2}{2} \log x - \dfrac{x^2}{4} \right )
I = \dfrac{x^2}{2} (\log x)^2 - \dfrac{x^2}{2} \log x + \dfrac{x^2}{4} + C
Final Answer: \boxed{ x \cdot (\log x)^2 = \dfrac{x^2}{2} (\log x)^2 - \dfrac{x^2}{2} \log x + \dfrac{x^2}{4} + C}
Question 15: Integrate (x^2 + 1) \cdot \log x
Solution:
Let, I = \int (x^2 + 1) \cdot \log x \ dx
Expand the integrand:
I = \int x^2 \cdot \log x + \log x \ dx
Integrate both terms separately:
First term:
I_1 = \int x^2 \cdot \log x \ dx
From Question 6:
I_1 = \dfrac{x^3}{3} \log x - \dfrac{x^3}{9}
Second term:
I_2 = \int \log x \ dx
Let’s use by parts for \int \log x \cdot dx:
Take u = \log x v = 1
Differentiate u: \dfrac{du}{dx} = \dfrac{1}{x}
Integrate v: \int 1 \ dx = x
Using by parts formula:
I_2 = \log x \cdot x - \int \dfrac{1}{x} \cdot x dx
I_2 = x \log x - \int dx
I_2 = x \log x - x
Now add both parts:
I = I_1 + I_2
I = \left( \dfrac{x^3}{3} \log x - \dfrac{x^3}{9} \right ) + (x \log x - x) + C
I = \dfrac{x^3}{3} \log x - \dfrac{x^3}{9} + x \log x - x + C
Final Answer: \boxed{\int (x^2 + 1) \cdot \log x \ dx = \left (\dfrac{x^3}{3} + x \right) \log x - \dfrac{x^3}{9} - x + C}
Question 16: Integrate e^x (\sin x + \cos x)
Solution:
Let I = \int e^x (\sin x + \cos x) dx
Let’s express as e^x [f(x) + f'(x)]:
Take f(x) = \sin x
f'(x) = \cos x
We know, \int e^x [f(x) + f'(x)] \ dx = e^x f(x) + C
so \int e^x [\sin x + \cos x] \ dx = e^x \sin x + C
Final Answer: \boxed{ \int e^x (\sin x + \cos x) \ dx = e^x \sin x + C}
Question 17: Integrate \dfrac{x \cdot e^x}{(1 + x)^2}
Solution:
Let I = \int \dfrac{x \cdot e^x}{(1 + x)^2} \ dx
Let’s try to write in the form e^x [f(x) + f'(x)]:
\dfrac{x}{(1 + x)^2} = \dfrac{(1 + x) - 1}{(1 + x)^2} = \dfrac{1}{1 + x} - \dfrac{1}{(1 + x)^2} =\dfrac{1}{1 + x} + \dfrac{-1}{(1 + x)^2}
So, I = \int e^x \left ( \dfrac{1}{1 + x} + \dfrac{-1}{(1 + x)^2} \right) \ dx
We know, \int e^x [f(x) + f'(x)] \ dx = e^x f(x) + C
Let f(x) = \dfrac{1}{1 + x}
Then, f'(x) = \dfrac{-1}{(1 + x)^2}
I = e^x \dfrac{1}{1 + x} + C
Final Answer: \boxed{\int \dfrac{x \cdot e^x}{(1 + x)^2} \ dx = \dfrac{e^x}{1 + x} + C}
Question 18: Integrate e^x \left( \dfrac{1 + \sin x}{1 + \cos x} \right)
Solution:
I = e^x \left( \dfrac{1 + \sin x}{1 + \cos x} \right)
Let’s try to write f(x) + f'(x) as numerator:
\left( \dfrac{1 + \sin x}{1 + \cos x} \right) = \left( \dfrac{1+ 2 \cdot \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \cdot \cos^2 \frac{x}{2}} \right)
\left( \dfrac{1 + \sin x}{1 + \cos x} \right) = \left( \dfrac{1}{2 \cdot \cos^2 \frac{x}{2}} + \dfrac{2 \cdot \sin \frac{x}{2} \cdot \cos \frac{x}{2}} {2 \cdot \cos^2 \frac{x}{2}} \right)
\left( \dfrac{1 + \sin x}{1 + \cos x} \right) = \dfrac{1}{2} \sec^2 \dfrac{x}{2} + \tan \dfrac{x}{2}
I = \int e^x \left( \dfrac{1}{2} \sec^2 \dfrac{x}{2} + \tan \dfrac{x}{2} \right) dx
Here
f(x) = \tan \dfrac{x}{2}
f'(x) = \sec^2 \dfrac{x}{2} \cdot \dfrac{1}{2}
We know, \int e^x [f(x) + f'(x)] \ dx = e^x f(x) + C
Thus, by the given formula:
I = e^x \tan \dfrac{x}{2} + C
Final Answer: \boxed{e^x \left( \dfrac{1 + \sin x}{1 + \cos x} \right) = e^x \tan \dfrac{x}{2} + C,}
Question 19: Integrate e^x \left( \dfrac{1}{x} - \dfrac{1}{x^2} \right)
Solution:
Let I = e^x \left( \dfrac{1}{x} - \dfrac{1}{x^2} \right)
Here
f(x) = \dfrac{1}{x}
f'(x) = -\dfrac{1}{x^2}
We know, \int e^x [f(x) + f'(x)] \ dx = e^x f(x) + C
Thus, by the given formula:
I = e^x \cdot \dfrac{1}{x} + C
Final Answer: \boxed{\int e^x \left( \dfrac{1}{x} - \dfrac{1}{x^2} \right ) dx = e^x \cdot \dfrac{1}{x} + C}
Question 20: Integrate \dfrac{(x-3) e^x}{(x-1)^3}
Solution:
Let I = \int \dfrac{(x-3) e^x}{(x-1)^3} \ dx
Let’s try to write in the form e^x [f(x) + f'(x)]
\dfrac{(x-3)}{(x-1)^3} = \dfrac{(x-1)-2}{(x-1)^3} = \dfrac{1}{(x-1)^2} + \dfrac{-2}{x-1^3}
I = \int e^x \left ( \dfrac{1}{(x-1)^2} + \dfrac{-2}{x-1^3} \right)
Here
f(x) = \dfrac{1}{(x-1)^2}
f'(x) = \dfrac{-2}{x-1^3}
We know, \int e^x [f(x) + f'(x)] \ dx = e^x f(x) + C
Thus, by the given formula:
I = e^x \left(\dfrac{1}{(x-1)^2} \right) + C
Final Answer: \boxed{ \int \dfrac{(x-3) e^x}{(x-1)^3} \ dx = \dfrac{e^x}{(x-1)^2} + C}
Question 21: Integrate e^{2x} \cdot \sin x
Solution:
Let I = \int e^{2x} \cdot \sin x \ dx
Let’s use integration by parts (twice):
Take u = \sin x v = e^{2x}
Differentiate u: \dfrac{du}{dx} = \cos x
Integrate v: \int e^{2x} , dx = \dfrac{1}{2} e^{2x}
By parts formula: \int u \cdot v \ dx = u \int v \ dx - \int \dfrac{du}{dx} \int v \ dx \cdot dx
Substituting values:
I = \sin x \cdot \dfrac{1}{2} e^{2x} - \int \cos x \cdot \dfrac{1}{2} e^{2x} dx
I = \dfrac{e^{2x}}{2} \sin x - \dfrac{1}{2} \int e^{2x} \cdot \cos x \ dx …….(1)
Now, let’s solve \int e^{2x} \cdot \cos x , dx using by parts again:
Take u = \cos x v = e^{2x}
Differentiate u: \dfrac{du}{dx} = -\sin x
Integrate v: \int v \ dx = \dfrac{1}{2} e^{2x}
Applying by parts:
\int e^{2x} \cdot \cos x \ dx = \cos x \cdot \dfrac{1}{2} e^{2x} - \int -\sin x \cdot \dfrac {1}{2} e^{2x} dx
\int e^{2x} \cdot \cos x \ dx = \dfrac{e^{2x}}{2} \cos x + \dfrac{1}{2} \int e^{2x} \cdot \sin x \ dx
Let’s call I = \int e^{2x} \cdot \sin x , dx.
\int e^{2x} \cdot \cos x \ dx = \dfrac{e^{2x}}{2} \cos x + \dfrac{1}{2} \cdot I
Substitute this result into the first equation:
I = \dfrac{e^{2x}}{2} \sin x - \dfrac{1}{2} \left( \dfrac{e^{2x}}{2} \cos x + \dfrac{1}{2} I \right)
I = \dfrac{e^{2x}}{2} \sin x - \dfrac{e^{2x}}{4} \cos x - \dfrac{1}{4} I
Bring like terms together:
I + \dfrac{1}{4} I = \dfrac{e^{2x}}{2} \sin x - \dfrac{e^{2x}}{4} \cos x
\dfrac{5}{4} I = \dfrac{e^{2x}}{2} \sin x - \dfrac{e^{2x}}{4} \cos x
I = \dfrac{4}{5} \left( \dfrac{e^{2x}}{2} \sin x - \dfrac{e^{2x}}{4} \cos x \right )
I = \dfrac{2}{5} e^{2x} \sin x - \dfrac{1}{5} e^{2x} \cos x + C
I = \dfrac{e^{2x}}{5} \cdot (2 \sin x - \cos x) + C
Final Answer: \boxed{ \int e^{2x} \cdot \sin x \ dx = \dfrac{e^{2x}}{5} \cdot (2 \sin x - \cos x) + C}
Question 22: Integrate \sin^{-1} \left( \dfrac{2x}{1 + x^2} \right)
Solution:
Let I = \int \sin^{-1} \left( \dfrac{2x}{1 + x^2} \right) dx
Let’s use substitution:
Let x = \tan \theta,
so dx = \sec^2 \theta , d\theta
Then,
\dfrac{2x}{1 + x^2} = \dfrac{2 \tan \theta}{1 + \tan^2 \theta} = \dfrac{2 \tan \theta}{\sec^2 \theta} = 2 \tan \theta \cdot \cos^2 \theta = \sin 2\theta
So,
\sin^{-1} \left( \dfrac{2x}{1 + x^2} \right) = 2\theta = 2 \tan^{-1} x
Therefore,
I = \int 2 \tan^{-1} x \ dx
Now, from earlier (Question 13):
\int \tan^{-1} x , dx = x \tan^{-1} x - \dfrac{1}{2} \log (1 + x^2)
So,
I = 2 \left[ x \tan^{-1} x - \dfrac{1}{2} \log (1 + x^2) \right ] + C
I = 2x \tan^{-1} x - \log (1 + x^2) + C
Final Answer: \boxed{\int \sin^{-1} \left( \dfrac{2x}{1 + x^2} \right) dx = 2x \tan^{-1} x - \log (1 + x^2) + C}
Question 23: \int x^2 e^{x^3} dx
A. \dfrac{1}{3} e^{x^3} + C
B. \dfrac{1}{2} e^{x^2} + C
C. \dfrac{1}{2} e^{x^3} + C
D. \dfrac{1}{3} e^{x^2} + C
Solution:
I = \int x^2 e^{x^3} dx
Let’s use substitution:
Let t = x^3 \implies dt = 3x^2 dx \implies x^2 dx = \dfrac{dt}{3}
So, I = \int e^{t} \cdot \dfrac{dt}{3}
I = \dfrac{1}{3} \int e^{t} dt
I = \dfrac{1}{3} e^{t} + C
I = \dfrac{1}{3} e^{x^3} + C
Correct answer is (A): \boxed{\int x^2 e^{x^3} dx= \dfrac{1}{3} e^{x^3} + C}
Question 24: \int e^x \cdot \sec x \cdot (1 + \tan x) \ dx
Solution:
Let I = \int e^x \cdot \sec x \cdot (1 + \tan x) \ dx
I = \int e^x \cdot ( \sec x + \sec x \tan x) \ dx
Here
f(x) = \sec x
f'(x) = \sec x \ \tan x
We know, \int e^x [f(x) + f'(x)] dx = e^x f(x) + C
Therefore,
I = e^x \cdot \sec x + C
Correct answer is (B): \boxed{\int e^x \cdot \sec x \cdot (1 + \tan x) \ dx = e^x \sec x + C }