The value of $\int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}} \frac{x \sin x^{2}}{\sin x^{2}+\sin \left(\ln 6-x^{2}\right)} d x$ is

(A) $\frac{1}{4} \ln \frac{3}{2}$

(B) $\frac{1}{2} \ln \frac{3}{2}$

(D) $\frac{1}{6} \ln \frac{3}{2}$

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Definite Integrals

The value of

\int_{l n 2}^{l n 3} \frac{x s i n x ^{2}}{s i n x ^{2} + s i n ( l n 6 - x ^{2} )} d x

Answer: (A)

Let

I = \int_{l n 2}^{l n 3} \frac{x s i n x ^{2}}{s i n x ^{2} + s i n ( l n 6 - x ^{2} )} d x

.

Substitute

x^{2} = t

, so

2 x d x = d t

, giving

I = \frac{1}{2} \int_{l n 2}^{l n 3} \frac{s i n t}{s i n t + s i n ( l n 6 - t )} d t

.

Using the property

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

, we also have

I = \frac{1}{2} \int_{l n 2}^{l n 3} \frac{s i n ( l n 6 - t )}{s i n ( l n 6 - t ) + s i n t} d t

.

Adding these two expressions:

2 I = \frac{1}{2} \int_{l n 2}^{l n 3} 1 d t = \frac{1}{2} (ln 3 - ln 2)

.

Therefore

I = \frac{1}{4} ln \frac{3}{2}

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Single correct option