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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Evaluate the following integrals:
$\int_{0}^\limits{{\frac{\pi}{4}}}(\tan\text{x}+\cot\text{x})^{-2}\text{ dx}$
✓

Answer

$\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}(\tan\text{x}+\cot\text{x})^{-2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\frac{1}{(\tan\text{x}+\cot\text{x})^{2}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\frac{1}{\Big(\frac{\sin^2\text{x}+\cos^2\text{x}}{\sin\text{x}\cos\text{x}}\Big)^2}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}(\sin\text{x}\cos\text{x})^2\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\sin^2\text{x}(1-\sin^2\text{x})\text{dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{{\frac{\pi}{4}}}\sin^2\text{x dx}-\int_{0}^\limits{{\frac{\pi}{4}}}\sin^4\text{x dx}$
We know that by reduction formula,
$\int\sin^{\text{n}}\text{x dx}=\frac{\text{n}-1}{\text{n}}\int\sin^{\text{n}-2}\text{x dx}-\frac{\cos\text{x}\sin^{\text{n}-1}}{\text{n}}$
For n = 2
$\Rightarrow\int\sin^2\text{x dx}=\frac{2-1}{2}\int1\text{ dx}-\frac{\cos\text{x}\sin\text{x}}{2}$
$\Rightarrow\int\sin^2\text{x dx}=\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}$
For n = 4
$\Rightarrow\int\sin^4\text{x dx}=\frac{4-1}{4}\int\sin^2\text{x dx}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
$\Rightarrow\int\sin^4\text{x dx}=\frac{3}{4}\Big\{\frac{1}{2}\times-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
Hence,
$\text{I}=\Big\{\frac{1}{2}\times-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}^{\frac{\pi}{4}}_0-\Big\{\frac{3}{4}\Big\{\frac{1}{2}\times-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}\Big\}^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\Big\{\frac{\pi}{8}-\frac{1}{4}\Big\}-\Big\{\frac{3}{4}\Big(\frac{\pi}{8}-\frac{1}{4}\Big)-\frac{1}{16}\Big\}$
$\Rightarrow\text{I}=\frac{\pi}{32}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}(\sin^{\text{x}}\cos\text{x})^2\text{dx}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}\sin^2\text{x}(1-\sin^2\text{x})\text{dx}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}\sin^2\text{x}-\sin^4\text{x dx}$
$\Rightarrow\int_{0}^\limits{\frac{\pi}{2}}\sin^2\text{x dx}-\int_{0}^\limits{\frac{\pi}{2}}\sin^4\text{x dx}$
We know, by reduction formula,
$\int\sin^{\text{n}}\text{x dx}=\frac{\text{n}-1}{\text{n}}\int\sin^{\text{n}-2}\text{x dx}-\frac{\cos\text{x }\sin^{\text{n}-1}\text{x}}{\text{n}}$
For n = 2
$\Rightarrow\int\sin^2\text{x dx}=\frac{2-1}{2}\int1\text{ dx}-\frac{\cos\text{x}\sin\text{x}}{2}$
$\Rightarrow\int\sin^2\text{x dx}=\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}$
For n = 4
$\Rightarrow\int\sin^4\text{x dx}=\frac{4-1}{4}\int\sin^2\text{x dx}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
$\Rightarrow\int\sin^4\text{x dx}=\frac{3}{4}\Big\{\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}$
Hence,
$\Big\{\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}^{\frac{\pi}{2}}_0-\Big\{\frac{3}{4}\Big\{\frac{1}{2}\text{x}-\frac{\cos\text{x}\sin\text{x}}{2}\Big\}-\frac{\cos\text{x}\sin^3\text{x}}{4}\Big\}^{\frac{\pi}{2}}_0$
$\Rightarrow\frac{\pi}{4}-\frac{3}{4}\Big\{\frac{\pi}{4}\Big\}$
$\Rightarrow\frac{\pi}{16}$

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