Question

integration of 1-sinx/cos^2x dx​

Answer 1

Step-by-step explanation:

Hope it will be helpful for u

Answer 2

The solution to the given integral is $\tan x- \sec x+ C$

Explanation:

To evaluate : $\int\dfrac{1-\sin x}{\cos^2x}\ dx$

This would become,

$\int(\dfrac{1}{\cos^2x}-\dfrac{\sin x}{\cos^2x})\ dx$

$=\int(\sec^x-\tan x\sec x )dx$  [$\because \sec x=\dfrac{1}{\cos x}\ \&\ \tan x=\dfrac{\sin x}{\cos x}$]

$=\int\sec^x\ dx-\int\tan x\sec x\ dx$

$=\tan x- \sec x+ C$ , where C is a constant.

[$\because\ \int \sec^x\ dx =\tan x$ and $\int\tan x\sec x\ dx=\sec x$]

Hence, the solution to the given integral is $\tan x- \sec x+ C$ .

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Integration of 3x^2dx + integration sinxdx =​

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