Evaluate: \(\int\cfrac{cosx}{(1+sin^2x)}dx\)∫ cos x/(1-sin2x)dx

1.	Evaluate: \(\int\cfrac{cosx}{(1+sin^2x)}dx\)∫ cos x/(1-sin2x)dx
Answer» To find: \(\int\cfrac{cosx}{(1+sin^2x)}dx\) Formula Used: \(\int\cfrac{dx}{a^2+x^2}=\cfrac{1}{a}tan^{-1}(\cfrac{x}{a})+C\) Let y = sin x … (1) Differentiating both sides, we get dy = cos x dx Substituting in given equation, \(\Rightarrow\)\(\int\cfrac{dy}{1+y^2}\) ⇒ tan^-1 y From (1), ⇒ tan^-1 (sin x) Therefore, \(\int\cfrac{cosx}{(1+sin^2x)}dx=tan^{-1}(sinx)+C\)

Answer»

To find: \(\int\cfrac{cosx}{(1+sin^2x)}dx\)

Formula Used: \(\int\cfrac{dx}{a^2+x^2}=\cfrac{1}{a}tan^{-1}(\cfrac{x}{a})+C\)

Let y = sin x … (1)

Differentiating both sides, we get

dy = cos x dx

Substituting in given equation,

\(\Rightarrow\)\(\int\cfrac{dy}{1+y^2}\)

⇒ tan^-1 y

From (1),

⇒ tan^-1 (sin x)

Therefore,

\(\int\cfrac{cosx}{(1+sin^2x)}dx=tan^{-1}(sinx)+C\)

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