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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int^\limits{\frac{\pi}{4}}_{0}\frac{\sin^2\text{x}\cos^2\text{x}}{(\sin^3\text{x}\cos^3\text{x})}\text{ dx}$
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Answer

Let $\text{I}=\int^\limits{\frac{\pi}{4}}_{0}\frac{\sin^2\text{x}\cos^2\text{x}}{(\sin^3\text{x}\cos^3\text{x})}\text{ dx}$$=\int^\limits{\frac{\pi}{4}}_{0}\frac{\sin^2\text{x}\cos^2\text{x}}{\cos^6\text{x}(\tan^3\text{x}+1)^2}\text{ dx}=\int^\limits{\frac{\pi}{4}}_{0}\frac{\tan^2\text{x}\sec^2\text{x}}{(\tan^3\text{x}+1)}\text{ dx}$
Put $\tan^3\text{x}+1=\text{z}$$\therefore\ 3\tan^{2}\text{x}\sec^2\text{x dx}=\text{dt}$
$\Rightarrow\tan^{2}\text{x}\sec^2\text{x dx}=\frac{\text{dz}}{3}$
When $\text{x}\rightarrow0,\text{z}\rightarrow1$ When $\text{x}\rightarrow\frac{\pi}{4},\text{z}\rightarrow2$$\therefore\ \text{I}=\frac{1}{3}\int^\limits{2}_1\frac{\text{dz}}{\text{z}^2}$
$=\frac{1}{3}\times-\Big[\frac{1}{\text{z}}\Big]^2_1$
$=-\frac{1}{3}\Big(\frac{1}{2}-1\Big)$
$=-\frac{1}{3}\times\Big(-\frac{1}{2}\Big)$
$=\frac{1}{6}$

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