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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int \frac{\sin^5\text{x}}{\cos^4\text{x}}\text{ dx}$
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Answer

$\int \frac{\sin^5\text{x}}{\cos^4\text{x}}\text{ dx}$ $=\int\Big(\frac{\sin^4\text{x}.\sin\text{x}}{\cos^4\text{x}}\Big)\text{dx}$ $=\int\frac{(\sin^2\text{x})^2\sin\text{x}}{\cos^4\text{x}}\text{ dx}$ $=\int\frac{(1-\cos^2\text{x})^2\sin\text{x}}{\cos^4\text{x}}\text{ dx}$ $=\int\Big(\frac{1+\cos^4\text{x}-2\cos^2\text{x}}{\cos^4\text{x}}\Big)\sin\text{x dx }$ $=\int\Big(\frac{1}{\cos^4\text{x}}+1-\frac{2}{\cos^2\text{x}}\Big)\sin\text{x dx}$ Let $\cos\text{x}=\text{t}$ $\Rightarrow-\sin\text{x}=\frac{\text{dt}}{\text{dx}}$ $$$\Rightarrow\sin\text{x dx}=-\text{dt}$Now, $=\int\Big(\frac{1}{\cos^4\text{x}}+1-\frac{2}{\cos^2\text{x}}\Big)\sin\text{x dx}$
$=-\int\big(\text{t}^{-4}+1-2\text{t}^{-2}\big)\text{ dt}$ $=-\Big[-\frac{\text{t}^{-4+1}}{-4+1}+\text{t}-\frac{2\text{t}^{-2+1}}{-2+1}\Big]+\text{C}$ $=-\Big[-\frac{1}{3\text{t}^3}+\text{t}+\frac{2}{\text{t}}\Big]+\text{C}$ $=\frac{1}{3\text{t}^3}-\text{t}-\frac{2}{\text{t}}+\text{C}$ $=\frac{1}{3\cos^3\text{x}}-\cos\text{x}-\frac{2}{\cos\text{x}}+\text{C}$

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