How to evaluate this indefinite integral $∫[10sin(x)]/[1+cos^2(x)]dx$ ?

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Answer 1 · 2017-12-02T07:15:16

The answer is $=-10arctan(cosx)+C$

Perform this integral by substitution

Let $u=cosx$ , $=>$ , $du=-sinxdx$

Therefore,

$int(10sinxdx)/(1+cos^2x)=-10int(du)/(1+u^2)$

Let $u=tantheta$

$du=sec^2thetad theta$

$1+tan^2theta=sec^2theta$

So,

$int(10sinxdx)/(1+cos^2x)=-10int(sec^2thetad theta)/(1+tan^2theta)$

$=-10int(sec^2thetad theta)/(sec^2theta)$

$=-10intd theta$

$=-10theta$

$=-10rctanu$

$=-10arctan(cosx)+C$

How to evaluate this indefinite integral ∫[10sin(x)]/[1+cos^2(x)]dx∫[10sin(x)]/[1+cos^2(x)]dx?