IIT-JEE 2011 · Paper 1 · Mathematics Q48

MathematicsSingle correct+3 / -1Question 48 of 69

The value of
$\int_{\sqrt{\ln2}}^{\sqrt{\ln3}}\frac{x\sin x^2}{\sin x^2+\sin\left(\ln6-x^2\right)}\,dx$
is

\dfrac14\ln\dfrac32

\dfrac12\ln\dfrac32

\ln\dfrac32

\dfrac16\ln\dfrac32

Answer and worked solution

Official answer: A

Substitute $u=x^2$ , $du=2x\,dx$ :
$I=\frac12\int_{\ln2}^{\ln3}\frac{\sin u}{\sin u+\sin(\ln6-u)}\,du.$

King's rule. Replacing $u$ by $\ln6-u$ swaps the two sines (the limits add to $\ln6$ ), so $2I=\frac12(\ln3-\ln2)$ and
$I=\frac14\ln\frac32.$

(A)

Topics:Properties of definite integrals Substitution

IIT-JEE 2011 · Paper 1, Mathematics question 48