Indefinite Integration 1 Question 22
23. The value of $\int _{\sqrt{\log 2}}^{\sqrt{\log 3}} \frac{x \sin x^{2}}{\sin x^{2}+\sin \left(\log 6-x^{2}\right)} d x$ is (2011)
(a) $\frac{1}{4} \log \frac{3}{2}$
(b) $\frac{1}{2} \log \frac{3}{2}$
(c) $\log \frac{3}{2}$
(d) $\frac{1}{6} \log \frac{3}{2}$
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Solution:
- Put $x^{2}=t \Rightarrow x d x=d t / 2$
$$ \therefore \quad I=\int _{\log 2}^{\log 3} \frac{\sin t \cdot \frac{d t}{2}}{\sin t+\sin (\log 6-t)} $$
Using, $\int _a^{b} f(x) d x=\int _a^{b} f(a+b-x) d x$
$$ \begin{aligned} & =\frac{1}{2} \int _{\log 2}^{\log 3} \frac{\sin (\log 2+\log 3-t)}{\sin (\log 2+\log 3-t)+\sin } d t \\ & =\frac{1}{2} \int _{\log 2}^{\log 3} \frac{\sin (\log 6-t)}{\sin (\log 6-t)+\sin (t)} d t \\ \therefore \quad I & =\int _{\log 2}^{\log 3} \frac{\sin (\log 6-t)}{\sin (\log 6-t)+\sin t} d t \end{aligned} $$
On adding Eqs. (i) and (ii), we get
$$ \begin{array}{rlrl} & & 2 I & =\frac{1}{2} \int _{\log 2}^{\log 3} \frac{\sin t+\sin (\log 6-t)}{\sin (\log 6-t)+\sin t} d t \\ \Rightarrow \quad 2 I & =\frac{1}{2}(t) _{\log 2}^{\log 3}=\frac{1}{2}(\log 3-\log 2) \\ \therefore \quad I & =\frac{1}{4} \log \frac{3}{2} \end{array} $$
