Q1. Find \displaystyle \int \sin^2(2x + 5) dx
Solution:
Recall, \sin^2 \theta = \frac{1 - \cos 2\theta}{2}
So,
\sin^2(2x + 5) = \frac{1 - \cos(4x + 10)}{2}
Therefore,
\int \sin^2(2x + 5) dx = \int \frac{1 - \cos(4x + 10)}{2} dx
\int \sin^2(2x + 5) dx = \frac{1}{2} \int dx - \frac{1}{2} \int \cos(4x + 10) dx
Integrate,
\int \sin^2(2x + 5) dx = \frac{1}{2}x - \frac{1}{2} \cdot \frac{\sin(4x + 10)}{4} + C
Final Answer: \boxed{ \int \sin^2(2x + 5) dx = \frac{1}{2}x - \frac{1}{8}\sin(4x + 10) + C }
Q2. Find \displaystyle \int \sin 3x \cos 4x dx
Solution:
Use product-to-sum:
\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]
So,
\sin 3x \cos 4x = \frac{1}{2}[\sin(7x) + \sin(-x)] = \frac{1}{2}[\sin 7x - \sin x]
Integrate,
\int \sin 3x \cos 4x dx = \frac{1}{2} \int [\sin 7x - \sin x] dx
\int \sin 3x \cos 4x dx = \frac{1}{2}\left(-\frac{\cos 7x}{7} + \frac{\cos x}{1}\right) + C
Final Answer: \boxed{ \int \sin 3x \cos 4x dx = -\frac{1}{14} \cos 7x + \frac{1}{2}\cos x + C }
Q3. Find \displaystyle \int \cos 2x \cos 4x \cos 6x dx
Solution:
First, use product-to-sum for two cosines:
\cos A \cos B = \frac{1}{2}[\cos(A + B) + \cos(A - B)]
So,
\cos 2x \cos 4x = \frac{1}{2} [\cos(2x - 4x) + \cos(2x + 4x)]
\cos 2x \cos 4x = \frac{1}{2} [\cos(-2x) + \cos(6x)]
\cos 2x \cos 4x = \frac{1}{2} [\cos 2x + \cos 6x]
Now,
\cos 2x \cos 4x \cos 6x = \frac{1}{2} [\cos 2x + \cos 6x] \cos 6x
\cos 2x \cos 4x \cos 6x = \frac{1}{2} [\cos 2x \cos 6x + \cos^2 6x]
Now expand both terms:
\cos 2x \cos 6x = \frac{1}{2} [\cos(2x - 6x) + \cos(2x + 6x)]
\cos 2x \cos 6x = \frac{1}{2} [\cos(-4x) + \cos(8x)]
\cos 2x \cos 6x = \frac{1}{2} [\cos 4x + \cos 8x]
\cos^2 6x = \frac{1 + \cos 12x}{2}
So,
\cos 2x \cos 4x \cos 6x = \frac{1}{2} \left( \frac{1}{2} [\cos 4x + \cos 8x] + \frac{1 + \cos 12x}{2} \right)
\cos 2x \cos 4x \cos 6x = \frac{1}{4} (\cos 4x + \cos 8x) + \frac{1}{4}(1 + \cos 12x)
\cos 2x \cos 4x \cos 6x = \frac{1}{4} \cos 4x + \frac{1}{4} \cos 8x + \frac{1}{4} + \frac{1}{4} \cos 12x
\int \cos 2x \cos 4x \cos 6x dx = \int \frac{1}{4} \cos 4x + \frac{1}{4} \cos 8x + \frac{1}{4} + \frac{1}{4} \cos 12x dx
Now integrate term by term:
\int \cos 2x \cos 4x \cos 6x dx = \frac{1}{4} \int \cos 4x; dx + \frac{1}{4} \int \cos 8x; dx + \frac{1}{4} \int dx + \frac{1}{4} \int \cos 12x; dx
\int \cos 2x \cos 4x \cos 6x dx= \frac{1}{4} \cdot \frac{\sin 4x}{4} + \frac{1}{4} \cdot \frac{\sin 8x}{8} + \frac{1}{4} x + \frac{1}{4} \cdot \frac{\sin 12x}{12} + C
Final Answer: \boxed{ \int \cos 2x \cos 4x \cos 6x dx= \frac{1}{4} [\frac{1}{4}\sin 4x + \frac{1}{8} \sin 8x + x + \frac{1}{12} \sin 12x] + C }
Q4. Find \displaystyle \int \sin^3 (2x + 1) dx
Solution:
Recall, \sin^3 \theta = \frac{3 \sin \theta - \sin 3\theta}{4}
So,
\sin^3 (2x + 1) = \frac{3 \sin(2x + 1) - \sin(6x + 3)}{4}
Integrate,
\int \sin^3 (2x + 1) dx = \frac{1}{4} \int 3 \sin(2x + 1) - \sin(6x + 3) dx
\int \sin^3 (2x + 1) dx = \frac{3}{4} \int \sin(2x + 1) dx - \frac{1}{4} \int \sin(6x + 3) dx
Using, \int \sin(ax + b); dx = -\frac{1}{a} \cos(ax + b) + C
So,
\int \sin^3 (2x + 1) dx = \frac{3}{4} \cdot \left(-\frac{1}{2} \cos(2x + 1)\right) - \frac{1}{4} \cdot \left(-\frac{1}{6} \cos(6x + 3)\right) + C
\boxed { \int \sin^3 (2x + 1) dx = -\frac{3}{8} \cos(2x + 1) + \frac{1}{24} \cos(6x + 3) + C }
We can also simplify this further
Using, \cos(3\theta) = 4\cos^3\theta - 3\cos\theta
So,
\cos(6x+3) = \cos(3\theta) = 4\cos^3(2x+1) - 3\cos(2x+1)
Substitute this into the answer:
\int \sin^3 (2x + 1) dx = -\frac{3}{8} \cos(2x+1) + \frac{1}{24} [4\cos^3(2x+1) - 3\cos(2x+1)] + C
\int \sin^3 (2x + 1) dx = -\frac{3}{8} \cos(2x+1) + \frac{1}{6} \cos^3(2x+1) - \frac{1}{8} \cos(2x+1)
\int \sin^3 (2x + 1) dx = -\frac{1}{2} \cos(2x+1) + \frac{1}{6} \cos^3(2x+1) + C
Final Answer: \boxed { \int \sin^3(2x+1),dx = -\frac{1}{2} \cos(2x+1) + \frac{1}{6} \cos^3(2x+1) + C}
Q5. Find \displaystyle \int \sin^3 x \cos^3 x; dx
Write: \sin^3 x \cos^3 x = \sin x (\sin^2 x) (\cos^3 x)
Recall: \sin^2 x = 1 - \cos^2 x.
So,
\sin^3 x \cos^3 x = \sin x (1 - \cos^2 x) \cos^3 x = \sin x \cos^3 x - \sin x \cos^5 x
\int \sin^3 x \cos^3 x dx = \int \sin x \cos^3 x dx - \int \sin x \cos^5 x dx
\int \sin x \cos^3 x dx
Let u = \cos x
Then, \frac{du}{dx} = -\sin x \implies \sin x dx = -du
Substituting in 1st part
\int \sin x \cos^3 x dx = \int \cos^3 x \sin x; dx = \int u^3 \cdot (-du) = -\int u^3 du = -\frac{u^4}{4} + C = -\frac{\cos^4 x}{4} + C
Substituting in 2nd part, using the same substitution:
\int \sin x \cos^5 x; dx = \int \cos^5 x \sin x; dx = \int u^5 \cdot (-du) = -\int u^5 du = -\frac{u^6}{6} + C = -\frac{\cos^6 x}{6} + C
Combine Results
Therefore,
\int \sin^3 x \cos^3 x dx = \left( -\frac{\cos^4 x}{4} \right) - \left( -\frac{\cos^6 x}{6} \right) + C
\int \sin^3 x \cos^3 x dx = -\frac{\cos^4 x}{4} + \frac{\cos^6 x}{6} + C
Final Answer: \boxed{ \int \sin^3 x \cos^3 x; dx = -\frac{1}{4} \cos^4 x + \frac{1}{6} \cos^6 x + C }
Q6. Find \displaystyle \int \sin x \sin 2x \sin 3x; dx
Solution:
We use product-to-sum formulas.
First,
\sin x \sin 2x = \frac{1}{2} [\cos(x - 2x) - \cos(x + 2x)] = \frac{1}{2} [\cos(-x) - \cos 3x] = \frac{1}{2} [\cos x - \cos 3x]
So,
\sin x \sin 2x \sin 3x = \frac{1}{2} [\cos x - \cos 3x] \sin 3x = \frac{1}{2} [\cos x \sin 3x - \cos 3x \sin 3x]
Now,
\cos x \sin 3x = \frac{1}{2} [\sin(3x + x) + \sin(3x - x)] = \frac{1}{2} [\sin 4x + \sin 2x]
\cos 3x \sin 3x = \frac{1}{2} \sin 6x
So,
\sin x \sin 2x \sin 3x = \frac{1}{2} \left( \frac{1}{2}[\sin 4x + \sin 2x] - \frac{1}{2} \sin 6x \right) = \frac{1}{4} (\sin 4x + \sin 2x - \sin 6x)
Integrate term by term:
\int \sin x \sin 2x \sin 3x dx = \frac{1}{4} \left( \int \sin 4x dx + \int \sin 2x dx - \int \sin 6x dx \right)
\int \sin x \sin 2x \sin 3x dx = \frac{1}{4} \left( -\frac{1}{4}\cos 4x - \frac{1}{2} \cos 2x + \frac{1}{6} \cos 6x \right) + C
Final Answer: \boxed {\int \sin x \sin 2x \sin 3x dx = \frac{1}{4} \left( -\frac{1}{4}\cos 4x - \frac{1}{2} \cos 2x + \frac{1}{6} \cos 6x \right) + C}
Q7. Find \int \sin 4x \sin 8x dx
Solution: Use the product-to-sum identity:
\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]
So
\sin 4x \sin 8x = \frac{1}{2}[\cos(4x) - \cos(12x)]
Now,
I = \int \sin 4x \sin 8x dx
I = \frac{1}{2}\int [\cos 4x - \cos 12x] dx
Integrating,
I = \frac{1}{2} \left(\frac{\sin 4x}{4} - \frac{\sin 12x}{12}\right) + C
I= \frac{1}{8} \sin 4x - \frac{1}{24} \sin 12x + C
Final Answer: \boxed{ I= \frac{1}{8} \sin 4x - \frac{1}{24} \sin 12x + C}
Q8. Find \int \frac{1 - \cos x}{1 + \cos x} dx
Solution:
Let’s simplify \frac{1 - \cos x}{1 + \cos x}:
Recall,
1 - \cos x = 2\sin^2 \frac{x}{2}
1 + \cos x = 2\cos^2 \frac{x}{2}
So,
\frac{1 - \cos x}{1 + \cos x} = \frac{2\sin^2 \frac{x}{2}}{2\cos^2 \frac{x}{2}} = \tan^2 \frac{x}{2}
Therefore,
I = \int \frac{1 - \cos x}{1 + \cos x} dx
I = \int \tan^2 \frac{x}{2} dx
Recall,
\tan^2 \theta = \sec^2 \theta - 1
So,
I = \int (\sec^2 \frac{x}{2} - 1) dx
I= \int \sec^2 \frac{x}{2} dx - \int dx
Let u = \frac{x}{2} \implies du = \frac{dx}{2} \implies dx = 2 du
So,
\int \sec^2 \frac{x}{2} dx = 2 \tan \frac{x}{2}
Therefore,
I = 2 \tan \frac{x}{2} - x + C
Final Answer: \boxed {\int \frac{1 - \cos x}{1 + \cos x} dx = 2 \tan \frac{x}{2} - x + C}
Q9. Find \int \frac{\cos x}{1 + \cos x} dx
Solution:
Let’s write 1 + \cos x = 2\cos^2 \frac{x}{2},
and \cos x = 1 - 2\sin^2 \frac{x}{2}
So,
\frac{\cos x}{1 + \cos x} = \frac{1 - 2\sin^2 \frac{x}{2}}{2\cos^2 \frac{x}{2}} = \frac{1}{2\cos^2 \frac{x}{2}} - \frac{2\sin^2 \frac{x}{2}}{2\cos^2 \frac{x}{2}}
= \frac{1}{2} \sec^2 \frac{x}{2} - \tan^2 \frac{x}{2}
So,
I = \int \frac{\cos x}{1 + \cos x} dx
I = \frac{1}{2} \int \sec^2 \frac{x}{2} dx - \int \tan^2 \frac{x}{2} dx
I = \frac{1}{2} \int \sec^2 \frac{x}{2} dx - \int (\sec^2 \frac{x}{2} - 1) dx
I = \frac{1}{2} \int \sec^2 \frac{x}{2} dx - \int \sec^2 \frac{x}{2} dx + \int 1 dx
I = \int 1 dx - \frac{1}{2} \int \sec^2 \frac{x}{2} dx
We know,
\int \sec^2 \frac{x}{2} dx = 2 \tan \frac{x}{2}
So, integrating
I = x -\frac{1}{2} \cdot 2 \tan \frac{x}{2} + C
I = x - \tan \frac{x}{2} + C
Final Answer: \boxed { \int \frac{\cos x}{1 + \cos x} dx = x - \tan \frac{x}{2} + C }
Q10. Find \int \sin^4 x dx
Solution: Recall,
\sin^4 x = \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x
So, I = \int \sin^4 x dx
I = \int \left( \frac{3}{8} - \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x \right) dx
I = \frac{3}{8}x - \frac{1}{2} \cdot \frac{\sin 2x}{2} + \frac{1}{8} \cdot \frac{\sin 4x}{4} + C
I = \frac{3}{8}x - \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C
Final Answer: \boxed{\int \sin^4 x dx = \frac{3}{8}x - \frac{1}{4}\sin 2x + \frac{1}{32}\sin 4x + C}
Q11. Find \int \cos^4 2x ; dx
Solution: Recall,
\cos^4 \theta = \frac{3}{8} + \frac{1}{2} \cos 2\theta + \frac{1}{8} \cos 4\theta
So for \theta = 2x,
\cos^4 2x = \frac{3}{8} + \frac{1}{2}\cos 4x + \frac{1}{8}\cos 8x
Therefore,
I = \int \cos^4 2x dx
I = \int \left(\frac{3}{8} + \frac{1}{2}\cos 4x + \frac{1}{8}\cos 8x\right) dx
I= \frac{3}{8}x + \frac{1}{2} \cdot \frac{\sin 4x}{4} + \frac{1}{8} \cdot \frac{\sin 8x}{8} + C
I = \frac{3}{8}x + \frac{1}{8}\sin 4x + \frac{1}{64}\sin 8x + C
Final Answer: \boxed { \int \cos^4 2x dx = \frac{3}{8}x + \frac{1}{8}\sin 4x + \frac{1}{64}\sin 8x + C }
Q12. Find \int \frac{\sin^2 x}{1 + \cos x} dx
Solution:
Recall: \sin^2 x = 1 - \cos^2 x,
but also, 1 + \cos x = 2\cos^2 \frac{x}{2} and \sin^2 x = 4\sin^2 \frac{x}{2} \cos^2 \frac{x}{2}
So,
\frac{\sin^2 x}{1 + \cos x} = \frac{4\sin^2 \frac{x}{2} \cos^2 \frac{x}{2}}{2\cos^2 \frac{x}{2}} = 2\sin^2 \frac{x}{2}
Therefore,
I = \int \frac{\sin^2 x}{1 + \cos x} dx
I = 2 \int \sin^2 \frac{x}{2} dx
Recall: \sin^2 \theta = \frac{1 - \cos 2\theta}{2}
So,
I = 2 \int \sin^2 \frac{x}{2} dx
I = 2 \int \frac{1 - \cos x}{2} dx
I = x - \sin x + C
\boxed{ \int \frac{\sin^2 x}{1 + \cos x} dx = x - \sin x + C}
Q13. Evaluate \displaystyle \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}; dx
Solution:
Recall \cos 2x = 2\cos^2 x - 1
Similarly \cos 2\alpha = 2\cos^2 \alpha - 1
So,
\cos 2x - \cos 2\alpha = [2\cos^2 x - 1] - [2\cos^2 \alpha - 1] = 2(\cos^2 x - \cos^2 \alpha)
Recall,
a^2 - b^2 = (a-b)(a+b)
Therefore,
\cos^2 x - \cos^2 \alpha = (\cos x - \cos \alpha)(\cos x + \cos \alpha)
Thus,
\cos 2x - \cos 2\alpha = 2(\cos x - \cos \alpha)(\cos x + \cos \alpha)
Now
I = \displaystyle \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}; dx
I = \int \frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} dx
I = \int 2(\cos x + \cos \alpha) dx
I= 2 \int \cos x ; dx + 2 \int \cos \alpha dx
Integrating,
I= 2 \sin x + 2x \cos \alpha + C
Final Answer: \boxed{\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} ; dx = 2 \sin x + 2x \cos \alpha + C }
Q14. Evaluate \displaystyle \int \frac{\cos x - \sin x}{1 + \sin 2x} dx
Solution:
We know:
1 = \sin^2 x + \cos^2 x
\sin 2x = 2\sin x \cos x
So,
1 + \sin 2x = \sin^2 x + \cos^2 x + 2\sin x \cos x = (\sin x + \cos x)^2
I = \int \frac{\cos x - \sin x}{1 + \sin 2x} dx
I = \int \frac{\cos x - \sin x}{(\sin x + \cos x)^2} ; dx
Use Substitution
Let u = \sin x + \cos x
Then,
\frac{du}{dx} = \cos x - \sin x \implies du = (\cos x - \sin x) dx
So,
(\cos x - \sin x) dx = du
Therefore,
I = \int \frac{du}{u^2}
Integrate
\int \frac{du}{u^2} = -\frac{1}{u} + C
Substitute back u = \sin x + \cos x
I = -\frac{1}{\sin x + \cos x} + C
Final Answer: \boxed{\int \frac{\cos x - \sin x}{1 + \sin 2x} ; dx = -\frac{1}{\sin x + \cos x} + C }
Q15. Evaluate \displaystyle \int \tan^3 2x \sec 2x dx
Solution:
Let t = \sec 2x
Differentiate both sides with respect to x:
\frac{dt}{dx} = 2\sec 2x \tan 2x
\frac{dt}{2} = \sec 2x \tan 2x dx
Write the Integral in Terms of t
I = \int \tan^3 2x \sec 2x dx
I = \int \tan^2 2x \tan 2x \sec 2x dx
I = \int \frac{\tan^2 2x}{2} dt
I = \frac{1}{2} \int (\sec^2 2x -1) dt
I = \frac{1}{2} \int (t^2 -1) dt
Integrating,
I = \frac{1}{2} \left( \frac{t^3}{3} - t \right) + C
I = \frac{1}{6} t^3 - \frac{1}{2} t + C
Recall, t = \sec 2x
I = \frac{1}{6} \sec^3 2x - \frac{1}{2} \sec 2x + C
Final Answer: \boxed{\int \tan^3 2x \sec 2x; dx = \frac{1}{6} \sec^3 2x - \frac{1}{2} \sec 2x + C}
Q16. Evaluate \displaystyle \int \tan^4 x dx
Solution:
Recall, \tan^4 x = (\tan^2 x)^2 = (\sec^2 x - 1)^2 = \sec^4 x - 2\sec^2 x + 1
So,
I = \int \tan^4 x dx
I = \int (\sec^4 x - 2\sec^2 x + 1) dx
I = \int \sec^4 x dx - 2 \int \sec^2 x .dx + \int 1.dx
I = \int (\sec^2 x) \cdot (\sec^2 x) dx - 2 \int \sec^2 x dx + \int 1.dx
I = \int (\sec^2 x) \cdot (1+ \tan^2 x)^2 dx - 2 \int \sec^2 x dx + \int 1.dx
I = \int (\sec^2 x) \cdot (1+ \tan^2 x) dx - 2 \int \sec^2 x dx + \int 1.dx
Let, t = \tan x
differentiating both sides with respect to x
\frac {dt}{dx} = \sec^2 x
dt = \sec^2 x.dx
Write the Integral in Terms of t,
I = \int (1+ t^2) dt - 2 \int \sec^2 x dx + \int 1.dx
Final Answer: \boxed{ \int \tan^4 x dx = \tan x + \frac{1}{3} \tan^3 x + x + C}
Q17. Evaluate \displaystyle \int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} dx
Solution:
I = \int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} dx
I = \int \frac{\sin^3 x}{\sin^2 x \cos^2 x} + \frac{\cos^3 x}{\sin^2 x \cos^2 x} dx
I = \int \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x} dx
I = \int \tan x.\sec x + \cot x.\cosec x . dx
Integrating,
I = \sec x - \cosec x + C
Final Answer: \boxed { \int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} dx = \sec x - \cosec x + C }
Q18. Evaluate \displaystyle \int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x} dx
Solution:
Recall, \cos 2x = 2\cos^2 x - 1
Also, \sin^2 x = 1 - \cos^2 x
So, \cos 2x + 2\sin^2 x = [2\cos^2 x - 1] + 2(1 - \cos^2 x) = 2\cos^2 x - 1 + 2 - 2\cos^2 x = 1
Therefore,
I = \int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x} dx
I = \int \frac{1}{\cos^2 x} dx
I = \int \sec^2 x dx
I = \tan x + C
Final Answer: \boxed{ \int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x} = \tan x + C }
Q19. Evaluate \displaystyle \int \frac{1}{\sin x \cos^3 x} dx
Solution:
I = \int \frac{1}{\sin x \cos^3 x} dx
multiplying numerator and denominator by \cos x ,
I = \int \frac{\cos x}{\sin x. \cos^4 x} dx
I = \int \frac{\sec^4x}{\tan x} dx
I = \int \frac{\sec^2x. \sec^2x}{\tan x} dx
I = \int \frac{(1+ \tan^2x) \sec^2x}{\tan x} dx
let t = \tan x
differentiating w.r.t x
\frac{dt}{dx} = \sec^2 x
dt = \sec^2 x.dx
Substituting in I
I = \int \frac{1+ t^2}{t} dt
I = \int \frac{1}{t}+ t.dt
Integrating,
I = \log(t) + \frac{t^2}{2} + C
Substituting back \tan x
I = \log|\tan x| + \frac{\tan^2 x}{2} + C
\boxed{ \int \frac{1}{\sin x \cos^3 x} dx = \log|\tan x| + \frac{\tan^2 x}{2} + C }
Q20. Evaluate \displaystyle \int \frac{\cos 2x}{(\cos x + \sin x)^2} dx
Solution:
Let’s write \cos 2x = \cos^2 x - \sin^2 x = (\cos x + \sin x)(\cos x - \sin x)
So,
\frac{\cos 2x}{(\cos x + \sin x)^2} = \frac{(\cos x + \sin x)(\cos x - \sin x)}{(\cos x + \sin x)^2} = \frac{\cos x - \sin x}{\cos x + \sin x}
So the integral becomes:
I = \int \frac{\cos 2x}{(\cos x + \sin x)^2} dx
I = \int \frac{\cos x - \sin x}{\cos x + \sin x} dx
Let u = \cos x + \sin x,
Then, du = -\sin x + \cos x dx = (\cos x - \sin x) dx
Therefore,
I = \int \frac{du}{u}
Integrating,
I = \ln |u| + C
I = \ln |\cos x + \sin x| + C
Final Answer: \boxed{\int \frac{\cos 2x}{(\cos x + \sin x)^2} dx = \ln |\cos x + \sin x| + C }
Q21. Evaluate \displaystyle \int \sin^{-1} (\cos x) dx
Solution:
Instead, use the identity:
\sin^{-1} (\cos x) = \frac{\pi}{2} - x \quad \text{for} \quad 0 \leq x \leq \frac{\pi}{2}
So,
I = \int \sin^{-1} (\cos x) dx
I = \int \left( \frac{\pi}{2} - x \right) dx
I = \frac{\pi}{2}x - \frac{x^2}{2} + C
Final Answer: \boxed{ \int \sin^{-1} (\cos x) dx = \frac{\pi}{2}x - \frac{x^2}{2} + C }
Question 22: \displaystyle \int \frac{1}{\cos(x-a)\cos(x-b)},dx
Solution:
I = \int \frac{1}{\cos(x-a)\cos(x-b)}dx
Multiply and Divide by sin(a−b)
I = \int \frac{\sin(a-b)}{\sin(a-b)\cos(x-a)\cos(x-b)}dx
I = \int \frac{\sin(a-b+x-x)}{\sin(a-b)\cos(x-a)\cos(x-b)}dx
I = \int \frac{\sin((x-b)+(a-x))}{\sin(a-b)\cos(x-a)\cos(x-b)}dx
Using: \sin(A-B) = \sin A \cos B - \cos A \sin B
I = \int \frac{\sin(x-b) \cos(a-x) - \cos(x-b) \sin(a-x))}{\sin(a-b)\cos(x-a)\cos(x-b)}dx
I = \int \frac{\sin(x-b) \cos(a-x)}{\sin(a-b)\cos(x-a)\cos(x-b)} - \frac{\cos((x-b) \sin(a-x)}{\sin(a-b)\cos(x-a)\cos(x-b)}dx
I = \int \frac{\sin(x-b) \cos(a-x)}{\sin(a-b)\cos(x-a)\cos(x-b)} - \frac{\cos((x-b) \sin(a-x)}{\sin(a-b)\cos(x-a)\cos(x-b)}dx
I = \int \frac{\tan(x-b) }{\sin(a-b)} - \frac{\tan((x-a) }{\sin(a-b)}dx
I = \frac{1}{\sin(a-b)} \int \tan(x-b) - \tan(x-a)dx
Integration,
I = \frac{1}{\sin(a-b)} [ - \log(\cos (x-b)) + \log(\cos (x-a)) + C
Final Answer: \boxed {\int \frac{1}{\cos(x-a)\cos(x-b)}dx = \frac{1}{\sin(a-b)} [\log(\frac{\cos (x-a)}{cos (x-b)}] + C }
Q23.Evaluate \displaystyle \int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} dx
Options:
(A) \tan x + \cot x + C
(B) \tan x + \csc x + C
(C) - \tan x + \cot x + C
(D) \tan x + \sec x + C
Solution:
Let’s split the numerator:
\frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x}
= \frac{\sin^2 x}{\sin^2 x \cos^2 x} - \frac{\cos^2 x}{\sin^2 x \cos^2 x}
= \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x}
= \sec^2 x - \csc^2 x
So the integral becomes:
I = \int (\sec^2 x - \csc^2 x) dx
I = \int \sec^2 x dx - \int \csc^2 x dx
Integrating,
I = \tan x + \cot x + C
Correct option: (A)
Q24. Evaluate \displaystyle \int \frac{e^x (1 + x)}{\cos^2 (x e^x)} dx
Options:
(A) -\cot (x e^x) + C
(B) \tan (x e^x) + C
(C) \tan (e^x) + C
(D) \cot (e^x) + C
Solution:
Let t = x e^x
Differentiate both sides:
\frac{dt}{dx} = e^x + x e^x = e^x (1 + x)
dt = e^x (1 + x) dx
Now substitute in the integral:
I = \int \frac{e^x (1 + x)}{\cos^2 (x e^x)} dx
I = \int \frac{1}{\cos^2 t} dt
I = \int \sec^2 t dt
I = \tan t + C
Substituting back, t = x e^x:
I = \tan (x e^x) + C
Correct option: (B)