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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\text{x}\sin\text{x}\cos\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\text{ dx}$
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Answer

Let $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\text{x}\sin\text{x}\cos\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\text{ dx}\ ...(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\big(\frac{\pi}{2}-\text{x}\big)\sin\big(\frac{\pi}{2}-\text{x}\big)\cos\big(\frac{\pi}{2}-\text{x}\big)}{\sin^4\big(\frac{\pi}{2}-\text{x}\big)+\cos^4\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\big(\frac{\pi}{2}-\text{x}\big)\cos\text{x}\sin\text{x}}{\cos^4\text{x}+\sin^4\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\big(\frac{\pi}{2}-\text{x}\big)\sin\text{x}\cos\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\Big(\text{x}+\frac{\pi}{2}-\text{x}\Big)\frac{\sin\text{x}\cos\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\text{ dx}$
$=\frac{\pi}{2}\int\limits^{\frac{\pi}{2}}_0\frac{\sin\text{x}\cos\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\text{ dx}$
Let $\sin^2\text{x}=\text{t},$ Then $2\sin\text{x}\cos\text{x dx}=\text{dt}$
When $\text{x}=0,\text{t}=0,\text{x}=\frac{\pi}{2},\text{t}=1$
Therefore,
$2\text{I}=\frac{\pi}{4}\int\limits^1_0\frac{\text{dt}}{\text{t}^2+(1-\text{t}^2)}$
$=\frac{\pi}{8}\int\limits^1_0\frac{\text{dt}}{\big(\text{t}-\frac{1}{2}\big)^2+\frac{1}{4}}$
$=\frac{\pi}{8}\times2\Big[\tan^{-1}(2\text{t}-1)\Big]^1_0$
$=\frac{\pi}{4}\Big(\frac{\pi}{4}+\frac{\pi}{4}\Big)$
Hence, $\text{I}=\frac{\pi^2}{16}$

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