Class 12 NCERT Maths – Chapter 7, Exercise 7.10 Solutions

Are you looking for NCERT Solutions for Class 12 Maths Integration Exercise 7.10? Then you have come to the right place.
This exercise covers questions based important properties of definite integrals that simplify evaluation of integrals in board exams.

Here, you’ll find stepwise solutions for Class 12, NCERT Maths Ex 7.10 explained in a clear, textbook style to help you score well in CBSE exams.
Before starting the exercise it is recommended that you go through the properties of definite integrals.

Question 1: Evaluate \int_{0}^{\frac{\pi}{2}} \cos^{2}x \ dx

Solution:
I = \int_{0}^{\frac{\pi}{2}} \cos^{2}x \ dx

I = \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos 2x}{2} \ dx

I = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 \ dx + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos 2x \ dx

I = \frac{1}{2} \cdot \frac{\pi}{2} + \frac{1}{2} \cdot \left[ \frac{\sin 2x}{2} \right]_{0}^{\frac{\pi}{2}}

I = \frac{\pi}{4} + \frac{1}{4}(0 - 0)

I = \frac{\pi}{4}

Final Answer: \boxed{\int_{0}^{\frac{\pi}{2}} \cos^{2}x \ dx = \frac{\pi}{4}}

Question 2: Evaluate \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \ dx

Solution:
I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \ dx … (1)

Using property P_{4}: \int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx with a = \frac{\pi}{2}:

I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin (\frac{\pi}{2} - x)}}{\sqrt{\sin (\frac{\pi}{2} - x)} + \sqrt{\cos (\frac{\pi}{2} - x)}} \ dx

I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \ dx … (2)

Adding (1) and (2)

2I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} \ dx

2I = \int_{0}^{\frac{\pi}{2}} 1 \ dx

2I = \Big [ x \Big]_{0}^{\frac{\pi}{2}}

2I = \dfrac{\pi}{2} - 0

2I = \dfrac{\pi}{2}

I = \frac{\pi}{4}

Final Answer: \boxed {\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \ dx = \frac{\pi}{4}}

Question 3: Evaluate \int_{0}^{\frac{\pi}{2}} \frac{\sin^{3/2}x}{\sin^{3/2}x + \cos^{3/2}x} , dx

Solution:
Let I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{3/2}x}{\sin^{3/2}x + \cos^{3/2}x} \ dx … (1)

Using property P_{4}: \int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx, with a = \frac{\pi}{2}:

I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{3/2} (\frac{\pi}{2} - x)}{\sin^{3/2} (\frac{\pi}{2} - x) + \cos^{3/2} (\frac{\pi}{2} - x)} \ dx

I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{3/2}x}{\sin^{3/2}x + \cos^{3/2}x} \ dx … (2)

Adding (1) and (2)

2I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{3/2}x + \sin^{3/2}x}{\sin^{3/2}x + \cos^{3/2}x} \ dx

2I = \int_{0}^{\frac{\pi}{2}} 1 \ dx

2I = \Big[x \Big]_{0}^{\frac{\pi}{2}}

2I = \dfrac{\pi}{2}

I = \dfrac{\pi}{4}

Final Answer: \boxed{ \int_{0}^{\frac{\pi}{2}} \frac{\sin^{3/2}x}{\sin^{3/2}x + \cos^{3/2}x} \ dx = \frac{\pi}{4}}

Question 4: Evaluate \int_{0}^{\frac{\pi}{2}} \frac{\cos^{5}x}{\sin^{5}x + \cos^{5}x} \ dx

Solution:
Let I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{5}x}{\sin^{5}x + \cos^{5}x} \ dx … (1)

Using property P_{4}: \int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx, with a = \frac{\pi}{2}:

Let I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^{5} (\frac{\pi}{2} - x)}{\sin^{5}(\frac{\pi}{2} - x) + \cos^{5}(\frac{\pi}{2} - x)} \ dx

I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{5}x}{\sin^{5}x + \cos^{5}x} \ dx … (2)

Adding (1) and (2)

2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{5}x + \cos^{5}x}{\sin^{5}x + \cos^{5}x} \ dx

2I = \int_{0}^{\frac{\pi}{2}} 1 / dx

2I = \Big[ x \Big]_{0}^{{\pi}{2}}

2I = \frac{\pi}{2}

I = \frac{\pi}{4}

Final Answer: \boxed {\int_{0}^{\frac{\pi}{2}} \frac{\cos^{5}x}{\sin^{5}x + \cos^{5}x} \ dx = \frac{\pi}{4} }

Question 5: Evaluate \int_{-5}^{5} |x+2| \ dx

Solution:
I = \int_{-5}^{5} |x+2| \ dx
Break at x = -2:

I = \int_{-5}^{-2} -(x+2) \ dx + \int_{-2}^{5} (x+2) \ dx

I = \int_{-2}^{-5} (x+2) \ dx + \int_{-2}^{5} (x+2) \ dx

I = \left[ \frac{x^{2}}{2} + 2x \right]_{-2}^{-5} + \left[ \frac{x^{2}}{2} + 2x \right]_{-2}^{5}

I = \left[ \frac{25}{2} - 10 \right] - \left[ 2 - 4 \right] + \left[ \frac{25}{2} + 10 \right] - \left[ 2 - 4 \right]

I = \frac{25}{2} - 10 - 2 + 4 + \frac{25}{2} + 10 - 2 + 4

I = \frac{50}{2} + 2 + 2

I = 25 + 4

I = 29

Final Answer: \boxed{ \int_{-5}^{5} |x+2| \ dx = 29}

Question 6: Evaluate \int_{2}^{8} |x-5| \ dx

Solution:
Let, I = \int_{2}^{8} |x-5| \ dx

Break at x=5:
I = \int_{2}^{5} -(x-5) \ dx + \int_{5}^{8} (x-5) \ dx

I = \int_{2}^{5} (5-x) \ dx + \int_{5}^{8} (x-5) \ dx

I = \left[ 5x - \frac{x^{2}}{2} \right]_{2}^{5} + \left[ \frac{x^{2}}{2} - 5x \right]_{5}^{8}

I = \left((25 - \frac{25}{2}) - (10 - 2) \right) + \left((32 - 40) - (\frac{25}{2} - 25) \right)

I = 25 - \frac{25}{2} - 8 - 8 - \frac{25}{2} + 25

I = 50 - 25 - 16

I = 9

Final Answer: \boxed{ \int_{2}^{8} |x-5| \ dx = 9 }

Question 7: Evaluate \int_{0}^{1} x(1-x)^{n} \ dx

Solution:
Let I = \int_{0}^{1} x(1-x)^{n} dx.

Put t = 1-x \implies dx = -dt.
limit x = 0 \implies t = 1 and x = 1 \implies t = 0

I = \int_{1}^{0} (1-t)t^{n}(-dt)

I = \int_{0}^{1} (1-t)t^{n} dt

So,
I = \int_{0}^{1} t^{n} dt - \int_{0}^{1} t^{n+1} dt

I = \Big[\frac{t^{n+1}}{n+1} \Big]_{0}^{1} - \Big[\frac{t^{n+2}}{n+2} \Big]_{0}^{1}

I = \frac{1}{n+1} - \frac{1}{n+2}

I = \frac{1}{(n+1)(n+2)}

Final Answer: \boxed{\int_{0}^{1} x(1-x)^{n} dx = \frac{1}{(n+1)(n+2)} }

Question 8: Evaluate \int_{0}^{\pi/4} \log(1+\tan x) \ dx

Solution:
Let I = \int_{0}^{\pi/4} \log(1+\tan x) \ dx.

Using property P_{4}: \int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx, with a = \frac{\pi}{4}:

I = \int_{0}^{\pi/4} \log(1+\tan(\pi/4 - x)) dx

I = \int_{0}^{\pi/4} \log\left(1+\frac{1-\tan x}{1+\tan x}\right) dx

I = \int_{0}^{\pi/4} \log\left(\frac{2}{1+\tan x}\right) dx

I = \int_{0}^{\pi/4} \log 2 \ dx - \int_{0}^{\pi/4} \log(1+\tan x) \ dx

2I = \frac{\pi}{4}\log 2

I = \frac{\pi}{8}\log 2

Final Answer: \boxed{ \int_{0}^{\pi/4} \log(1+\tan x) \ dx = \frac{\pi}{8}\log 2 }

Question 9: Evaluate \int_{0}^{2} x\sqrt{2-x} \ dx

Solution:
I = \int_{0}^{2} x\sqrt{2-x} \ dx

Let t = 2-x \implies dx = -dt

When x=0, t=2; when x=2, t=0.

I = \int_{2}^{0} (2-t)\sqrt{t}(-dt)

I= \int_{0}^{2} (2-t)\sqrt{t} \ dt

I = \int_{0}^{2} 2\sqrt{t} dt - \int_{0}^{2} t^{3/2} dt

I = \big[ \frac{2 \cdot 2 \cdot t^{3/2}}{3} \big]_{0}^{2} - \Big[ \frac{2 \cdot t^{5/2}}{5} \big]_{0}^{2}

I = \frac{4}{3} \cdot 2^{3/2} - \frac{2}{5} \cdot 2^{5/2}

I = \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5}

I = \frac{40\sqrt{2} - 24\sqrt{2}}{15}

I = \boxed{\frac{16\sqrt{2}}{15}}

Final Answer: \boxed{ \int_{0}^{2} x\sqrt{2-x} \ dx = \frac{16\sqrt{2}}{15}}

Question 10: Evaluate \int_{0}^{\pi/2} (2\log \sin x - \log \sin 2x) \ dx

Solution:
I = \int_{0}^{\pi/2} 2\log \sin x \ dx - \int_{0}^{\pi/2} \log \sin 2x \ dx

I = \int_{0}^{\pi/2} 2\log \sin x \ dx - \int_{0}^{\pi/2} \log (2\sin x \cos x) \ dx

I = \int_{0}^{\pi/2} 2\log \sin x \ dx - ( \int_{0}^{\pi/2} \log (2) \ dx + \int_{0}^{\pi/2} \log \sin x \ dx + \int_{0}^{\pi/2} \log \cos x \ dx )

I = \int_{0}^{\pi/2} 2\log \sin x \ dx - \int_{0}^{\pi/2} \log (2) \ dx - \int_{0}^{\pi/2} \log \sin x \ dx - \int_{0}^{\pi/2} \log \cos x \ dx

In the last integral Using property P_{4}: \int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx, with a = \frac{\pi}{2}:

I = \int_{0}^{\pi/2} 2\log \sin x \ dx - \int_{0}^{\pi/2} \log (2) \ dx - \int_{0}^{\pi/2} \log \sin x \ dx - \int_{0}^{\pi/2} \log \cos (x - \frac{\pi}{2}) \ dx

I = \int_{0}^{\pi/2} 2\log \sin x \ dx - \int_{0}^{\pi/2} \log (2) \ dx - \int_{0}^{\pi/2} \log \sin x \ dx - \int_{0}^{\pi/2} \log \sin x \ dx

I = - \int_{0}^{\pi/2} \log (2) \ dx

I = - \big[x \cdot \log (2) \big]_{0}^{\pi/2}

I = - \log(2) \big[\frac{\pi}{2} - 0]

I = - \dfrac{\pi}{2} \log(2)

Final Answer: \boxed{\int_{0}^{\pi/2} (2\log \sin x - \log \sin 2x) \ dx = -\frac{\pi}{2} \log 2}

Question 11: Evaluate \int_{-\pi/2}^{\pi/2} \sin^{2}x \ dx

Solution:
Let I = \int_{-\pi/2}^{\pi/2} \sin^{2}x \ dx

Since, \sin^{2}x \text{ is even, }

I = 2\int_{0}^{\pi/2}\sin^{2}x \ dx

Since, \sin^{2}x = \frac{1+\cos 2x}{2}

I = 2\int_{0}^{\pi/2} \frac{1+\cos 2x}{2} \ dx

I = 2\left[\frac{1}{2}\int_{0}^{\pi/2}1 \ dx + \frac{1}{2}\int_{0}^{\pi/2}\cos 2x \ dx\right]

I =\int_{0}^{\pi/2}1 \ dx + \frac{1}{2}\int_{0}^{\pi/2}\cos 2x \ dx

I = \big[ x \big]_{0}^{\pi/2} + \big[ \frac{\sin 2x}{2} \big]_{0}^{\pi/2}

I = \big[ \frac{\pi}{2} - 0 \big] + \big[ \frac{\sin {\pi} - \sin {0} }{2} \big]

I = \frac{\pi}{2}

Final Answer: \boxed{ \int_{-\pi/2}^{\pi/2} \sin^{2}x \ dx = \frac{\pi}{2}}

Question 12: Evaluate \int_{0}^{\pi}\frac{x}{1+\sin x},dx

Solution:
I = \int_{0}^{\pi}\frac{x}{1+\sin x} dx … (1)

Using property P_{4}: \int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx, with a = \pi:

I = \int_{0}^{\pi}\frac{\pi-x}{1+\sin (\pi - x)} \ dx

I = \int_{0}^{\pi}\frac{\pi-x}{1+\sin x} \ dx … (2)

Adding (1) and (2)

2I = \int_{0}^{\pi}\frac{x}{1+\sin x} + \frac{\pi-x}{1+\sin x} dx

2I = \int_{0}^{\pi}\frac{\pi}{1+\sin x} \ dx

2I = \pi\int_{0}^{\pi}\frac{1}{1+\sin x} \ dx

Now, \frac{1}{1+\sin x} = \frac{1-\sin x}{\cos^{2}x} = \sec^{2}x - \sec x \tan x

2I = \pi \int_{0}^{\pi}\sec^{2}x - \sec x \tan x \ dx

2I = \pi \bigl(\tan x - \sec x\bigr)_{0}^{\pi}

2I = \pi \big[ \bigl(\tan \pi - \sec \pi \bigr) - \bigl(\tan 0 - \sec 0 \bigr) \big]

2I = \pi \big[ \bigl( 0 + 1 \bigr) - \bigl(0 - 1 \bigr) \big]

2I = 2 \pi

I = \pi

Final Answer: \boxed { \int_{0}^{\pi}\frac{x}{1+\sin x} dx = \pi }

Question 13: Evaluate \int_{-\pi/2}^{\pi/2} \sin^{7}x \ dx

Solution:
\sin^{7}x is an odd function.

We know: P_5: \int_{-a}^{a} f(x) \ dx = 0, if f is an odd function.
i.e. Integral of odd function over symmetric limits is zero.

Final Answer \boxed{\int_{-\pi/2}^{\pi/2} \sin^{7}x = 0}

Question 14: Evaluate \int_{0}^{2\pi} \cos^{5}x \ dx

Solution:
I = \int_{0}^{2\pi} \cos^{5}x \ dx

I = \int_{0}^{2\pi} \cos^{4}x \cdot \cos x \ dx

I = \int_{0}^{2\pi} (\cos^{2}x)^2 \cdot \cos x \ dx

I = \int_{0}^{2\pi} (1 - \sin^{2}x)^2 \cdot \cos x \ dx

Let \sin x = t \implies \cos x = \frac{dt}{dx}
when x = 0, t = 0 , when x = 2\pi, t = 0

I = \int_{0}^{0} (1 - t^2)^2 \ dt

I = 0

Final Answer: \boxed{\int_{0}^{2\pi} \cos^{5}x \ dx = 0}

Question 15: Evaluate \int_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \ dx

Solution:
I = \int_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \ dx

Using property P_{4}: \int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx, with a = \frac{\pi}{2}:

I = \int_{0}^{\pi/2} \frac{\sin (\frac{\pi}{2} - x) - \cos (\frac{\pi}{2} - x)}{1 + \sin (\frac{\pi}{2} - x) \cos (\frac{\pi}{2} - x) } \ dx

I = \int_{0}^{\pi/2} \frac{\cos x - \sin x}{1 + \cos x \sin x \ } dx

I = \int_{0}^{\pi/2} - \frac{\sin x - \cos x}{1 + \sin x \cos x \ } dx

So, I = -I

2I = 0

I = 0

Final Answer: \boxed{ \int_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \ dx = 0 }

Question 16: Evaluate \int_{0}^{\pi} \log(1+\cos x) \ dx

Solution:
I = \int_{0}^{\pi} \log(1+\cos x) \ dx … (1)

Using property P_{4}: \int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx, with a = \pi :

I = \int_{0}^{\pi} \log(1+\cos (\pi - x)) \ dx

I = \int_{0}^{\pi} \log(1-\cos x) \ dx … (2)

Adding (1) and (2)

2I = \int_{0}^{\pi} \log(1-\cos^2 x) \ dx

2I = \int_{0}^{\pi} \log(\sin^2 x) \ dx

2I = 2 \cdot \int_{0}^{\pi} \log(\sin x) \ dx

I = \int_{0}^{\pi} \log(\sin x) \ dx

Using Property P₆: \int_{0}^{2a} f(x) \ dx = 2 \int_{0}^{a} f(x) \ dx, \ \ \ \ \text{if } f(2a-x) = f(x)

I = 2 \cdot \int_{0}^{\pi/2} \log(\sin x) \ dx … (3)

Using property P_{4}: \int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx, with a = \frac{\pi}{2} :

I = 2 \cdot \int_{0}^{\pi/2} \log(\sin (\frac{\pi}{2} - x)) \ dx

I = 2 \cdot \int_{0}^{\pi/2} \log(\cos x) \ dx … (4)

Adding (3) and (4)

2I = 2 \big[ \int_{0}^{\pi/2} \log(\sin x) \ dx + \int_{0}^{\pi/2} \log(\cos x) \ dx \big]

2I = 2 \big[ \int_{0}^{\pi/2} \log(\sin x \cos x ) \ dx \big]

I = \int_{0}^{\pi/2} \log(\sin x \cos x ) \ dx

I = \int_{0}^{\pi/2} \log(\frac{2}{2}\sin x \cos x ) \ dx

I = \int_{0}^{\pi/2} \log (\sin{2x}) - \log{2} \ dx

I = \int_{0}^{\pi/2} \log (\sin{2x}) \ dx - \int_{0}^{\pi/2} \log{2} \ dx

For the 1st integral let,
2x = t \implies 2dx = dt \implies dx = \frac{dt}{2}
when, x = 0, t = 0 when, x = \frac{\pi}{2}, t = \pi

I = \frac{1}{2}\int_{0}^{\pi} \log (\sin{t}) \ dt - \int_{0}^{\pi/2} \log{2} \ dx

Using Property P₀: \int_a^b f(t),dt = \int_a^b f(x),dx

I = \frac{1}{2}\int_{0}^{\pi} \log (\sin{x}) \ dx - \int_{0}^{\pi/2} \log{2} \ dx

I = \frac{1}{2} \cdot I - \int_{0}^{\pi/2} \log{2} \ dx

\frac{1}{2} \cdot I = - \int_{0}^{\pi/2} \log{2} \ dx

\frac{1}{2} \cdot I = - \log{2} \big[ x \big]_{0}^{\pi/2}

\frac{1}{2} \cdot I = - \log{2} \big[ \frac{\pi}{2} - 0 \big]

\frac{1}{2} \cdot I = - \log{2} \frac{\pi}{2}

I = - \pi \log{2}

Final Answer: \boxed { \int_{0}^{\pi} \log(1+\cos x) \ dx = - \pi \log{2}}

Question 17: Evaluate \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} \ dx

Solution:
I = \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} \ dx … (1)

Using property P_{4}: \int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx, with a = a:

I = \int_{0}^{a} \frac{\sqrt{a- x}}{\sqrt{a - x}+\sqrt{a- (a- x}} \ dx

I = \int_{0}^{a} \frac{\sqrt{a- x}}{\sqrt{a - x}+\sqrt{x}} \ dx … (2)

Adding (1) and (2)

2I = \int_{0}^{a}1 \ dx

2I = \big[ x \big]_{0}^{a}

2I = a

I = \frac{a}{2}

Final Answer: \boxed{\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} \ dx = \frac{a}{2} }

Question 18: Evaluate \int_{0}^{4} |x-1| \ dx

Solution:
I = \int_{0}^{4} |x-1| \ dx

I = \int_{0}^{1}(1-x) \ dx + \int_{1}^{4}(x-1) \ dx

I = \left[x-\tfrac{x^{2}}{2}\right]_{0}^{1} + \left[\tfrac{x^{2}}{2}-x\right]_{1}^{4}

I = \tfrac{1}{2} + \tfrac{9}{2}

I = 5

Final Answer: \boxed{ \int_{0}^{4} |x-1| \ dx = 5}

Question 19: Show that \int_{0}^{a} f(x)g(x),dx = 2\int_{0}^{a} f(x),dx, if f and g are defined as f(x)=f(a-x) and g(x)+g(a-x)=4.

Solution:
I = \int_{0}^{a} f(x)g(x) \ dx … (1)

Using property P_{4}: \int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx, with a = a:

I = \int_{0}^{a} f(a-x)g(a-x) \ dx

Given: f(x)=f(a-x)

I = \int_{0}^{a} f(x)g(a-x) \ dx … (2)

Adding (1) and (2)
2I = \int_{0}^{a} f(x)\big(g(x)+g(a-x)\big) \ dx

Given: g(x)+g(a-x)=4

2I = \int_{0}^{a} f(x)\cdot 4 \ dx

2I = 4\int_{0}^{a} f(x),dx

I = 2\int_{0}^{a} f(x),dx

Hence Proved: \boxed{\int_{0}^{a} f(x)g(x) \ dx = 2\int_{0}^{a} f(x) \ dx}

Question 20: Evaluate \int_{-\pi/2}^{\pi/2}(x^{3}+x\cos x+\tan^{5}x+1) \ dx

(A) 0

(B) 2

(C) \pi

(D) 1

Solution:
I = \int_{-\pi/2}^{\pi/2}(x^{3}+x\cos x+\tan^{5}x+1) \ dx

I = \int_{-\pi/2}^{\pi/2} x^{3} \ dx + \int_{-\pi/2}^{\pi/2} x\cos x \ dx + \int_{-\pi/2}^{\pi/2}\tan^{5}x \ dx + \int_{-\pi/2}^{\pi/2} 1 \ dx

x^3 , x\cos x \text{and} \tan^{5}x are all functions

We know: P_5: \int_{-a}^{a} f(x) \ dx = 0, if f is an odd function.

Odd parts vanish:

\int_{-\pi/2}^{\pi/2}x^{3} \ dx = 0,

\int_{-\pi/2}^{\pi/2}x\cos x \ dx = 0,

\int_{-\pi/2}^{\pi/2}\tan^{5}x \ dx = 0

Only constant remains:

I = 0 + 0 + 0 + \int_{-\pi/2}^{\pi/2}1 \ dx

I = \big[ x \big]_{-\pi/2}^{\pi/2}

I = \frac{\pi}{2} - \frac{-\pi}{2}

I = \pi

\boxed{ \int_{-\pi/2}^{\pi/2}(x^{3}+x\cos x+\tan^{5}x+1) \ dx = \pi}

Correct Option: (C)

Question 21: Evaluate \int_{0}^{\pi/2}\log\left(\frac{4+3\sin x}{4+3\cos x}\right) \ dx

(A) 0

(B) 2

(C) \pi

(D) 1

Solution:
I = \int_{0}^{\pi/2}\log\left(\frac{4+3\sin x}{4+3\cos x}\right) \ dx

Using property P_{4}: \int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx, with a = \frac{\pi}{2}:

I = \int_{0}^{\pi/2}\log\left(\frac{4+3\sin (\tfrac{\pi}{2} - x)}{4+3\cos (\tfrac{\pi}{2} - x)}\right) \ dx

I = \int_{0}^{\pi/2}\log\left(\frac{4+3\cos x}{4+3\sin x}\right) \ dx

I = - \int_{0}^{\pi/2}\log\left(\frac{4+3\sin x}{4+3\cos x}\right) \ dx

I = -I

2I = 0

I = 0

\boxed{\int_{0}^{\pi/2}\log\left(\frac{4+3\sin x}{4+3\cos x}\right) \ dx = 0}

Correct Option: (A)