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Question
Write a value of\[\int\frac{\sin x}{\cos^3 x} \text{ dx }\]
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Solution
\[\text{ Let I }= \int \frac{\sin x}{\cos^3 x}dx\]
\[\text{ Let cos x }= t\]
\[ \Rightarrow - \text{ sin x dx} = dt\]
\[ \Rightarrow \text{ sin x dx }= - dt\]
\[ \therefore I = - \int \frac{dt}{t^3}\]
\[ = - \int t^{- 3} dt\]
\[ = - \left[ \frac{t^{- 3 + 1}}{- 3 + 1} \right] + C\]
\[ = \frac{1}{2 t^2} + C\]
\[ = \frac{1}{2 \cos^2 x} + C \left( \because t = \cos x \right)\]
\[ = \frac{1}{2} \text{ sec}^2 x + C\]
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