Prove that $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$ and hence evaluate $\int_{0}^{\pi / 2} \frac{x}{\sin x+\cos x} d x$.

Property proved; $\int_{0}^{\pi/2}\frac{x}{\sin x+\cos x}\,dx=\frac{\pi}{2 \sqrt{2}}\log (\sqrt{2}+1)$

Definite Integrals — CBSE Class 12 · 2019 Board Question with Solution

Prove that

\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x

and hence evaluate

\int_{0}^{π /2} \frac{x}{s i n x + c o s x} d x

Answer: Property proved; $\int_{0}^{π /2} \frac{x}{s i n x + c o s x} d x = \frac{π}{2 2} lo g (2 + 1)$

Proof: in the RHS put

t = a - x \Rightarrow d x = - d t

;

x = 0 \to t = a

x = a \to t = 0

. Then

\int_{0}^{a} f (a - x) d x = - \int_{a}^{0} f (t) d t = \int_{0}^{a} f (t) d t = \int_{0}^{a} f (x) d x .

Hence the property holds.
Now let

I = \int_{0}^{π /2} \frac{x}{sin x + cos x} d x

. Apply the property with

a = \frac{π}{2}

(so

sin \leftrightarrow cos

I = \int_{0}^{π /2} \frac{\frac{π}{2} - x}{cos x + sin x} d x \Rightarrow 2 I = \frac{π}{2} \int_{0}^{π /2} \frac{d x}{sin x + cos x}

Write

sin x + cos x = 2 sin (x + \frac{π}{4})

2 I = \frac{π}{2 2} \int_{0}^{π /2} cosec (x + \frac{π}{4}) d x = \frac{π}{2 2} [lo g tan (\frac{x}{2} + \frac{π}{8})]_{0}^{π /2}

= \frac{π}{2 2} [lo g tan \frac{3 π}{8} - lo g tan \frac{π}{8}] = \frac{π}{2 2} lo g \frac{2 + 1}{2 - 1} = \frac{π}{2} lo g (2 + 1)

Hence

I = \frac{π}{2 2} lo g (2 + 1)

CBSE Class 12 Board · 2019 · Subjective