Evaluate: \(\int\cfrac{sinx}{(1+cos^2x)}dx\)∫ sin x/(1+cos2x)dx

1.	Evaluate: \(\int\cfrac{sinx}{(1+cos^2x)}dx\)∫ sin x/(1+cos2x)dx
Answer» To find: \(\int\cfrac{sinx}{(1+cos^2x)}dx\) Formula Used: \(\int\cfrac{dx}{a^2+x^2}=\cfrac{1}{a}tan^{-1}(\cfrac{x}{a})+c\) Let y = cos x … (1) Differentiating both sides, we get dy = –sin x dx Substituting in given equation, \(\Rightarrow\)\(\int\cfrac{-dy}{1+y^2}\) ⇒ – tan -1 y From (1), ⇒ –tan -1 (cos x) Therefore, \(\int\cfrac{sinx}{1+cos^2x}dx=-tan^{-1}{(cosx)}+C\)

Answer»

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

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

Let y = cos x … (1)

Differentiating both sides, we get

dy = –sin x dx

Substituting in given equation,

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

⇒ – tan -1 y

From (1),

⇒ –tan -1 (cos x)

Therefore,

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

Discussion