Find the following integral:int frac{1-sin x} | vedclass

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Question

(A) We have the integral:$\int \frac{1-\sin x}{\cos ^{2} x} d x$Split the fraction into two parts:$= \int \left( \frac{1}{\cos ^{2} x} - \frac{\sin x}

Vedclass · Accepted Answer

$	an x - \sec x + C$

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(A) We have the integral:$\int \frac{1-\sin x}{\cos ^{2} x} d x$Split the fraction into two parts:$= \int \left( \frac{1}{\cos ^{2} x} - \frac{\sin x}{\cos ^{2} x} ight) d x$$= \int \sec ^{2} x d x - \int \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} d x$$= \int \sec ^{2} x d x - \int 	an x \sec x d x$Using the standard integration formulas $\int \sec ^{2} x d x = 	an x + C_1$ and $\int \sec x 	an x d x = \sec x + C_2$:$= 	an x - \sec x + C$

Find the following integral:
$\int \frac{1-\sin x}{\cos ^{2} x} d x$

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Find the following integral:$\int \frac{1-\sin x}{\cos ^{2} x} d x$