SATHEE: Indefinite Integration 1 Question 22

Indefinite Integration 1 Question 22

23. The value of $\int_{\sqrt{\log 2}}^{\sqrt{\log 3}} \frac{x \sin x^{2}}{\sin x^{2} + \sin (\log 6 - x^{2})} d x$ is (2011)

(a) $\frac{1}{4} \log \frac{3}{2}$

(b) $\frac{1}{2} \log \frac{3}{2}$

(d) $\frac{1}{6} \log \frac{3}{2}$

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Solution:

Put $x^{2} = t \Rightarrow x d x = d t / 2$

$∴ I = \int_{\log 2}^{\log 3} \frac{\sin t \cdot \frac{d t}{2}}{\sin t + \sin (\log 6 - t)}$

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

$\begin{aligned} = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 2 + \log 3 - t)}{\sin (\log 2 + \log 3 - t) + \sin} d t \\ = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 6 - t)}{\sin (\log 6 - t) + \sin (t)} d t \\ ∴ I & = \int_{\log 2}^{\log 3} \frac{\sin (\log 6 - t)}{\sin (\log 6 - t) + \sin t} d t \end{aligned}$

On adding Eqs. (i) and (ii), we get

$\begin{aligned} 2 I & = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t + \sin (\log 6 - t)}{\sin (\log 6 - t) + \sin t} d t \\ \Rightarrow 2 I & = \frac{1}{2} (t)_{\log 2}^{\log 3} = \frac{1}{2} (\log 3 - \log 2) \\ ∴ I & = \frac{1}{4} \log \frac{3}{2} \end{aligned}$

Indefinite Integration 1 Question 22

23. The value of $\int_{\sqrt{\log 2}}^{\sqrt{\log 3}} \frac{x \sin x^{2}}{\sin x^{2} + \sin (\log 6 - x^{2})} d x$ is (2011)

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Indefinite Integration 1 Question 22

23. The value of ∫log⁡2log⁡3xsin⁡x2sin⁡x2+sin⁡(log⁡6−x2)dx is (2011)

Solution:

Engineering Preparation

Medical Preparation

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23. The value of $\int_{\sqrt{\log 2}}^{\sqrt{\log 3}} \frac{x \sin x^{2}}{\sin x^{2} + \sin (\log 6 - x^{2})} d x$ is (2011)