int {{e^{{{cos }^2}x}}sin 2x,dx = } | vedclass

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Question

(B) Let $I = \int {{e^{{{\cos }^2}x}}\sin 2x\,dx}$.Substitute $t = {\cos ^2}x$.Then,differentiating both sides with respect to $x$,we get $dt = 2\cos

Vedclass · Accepted Answer

$-{e^{{{\cos }^2}x}} + c$

Vedclass · Answer

(B) Let $I = \int {{e^{{{\cos }^2}x}}\sin 2x\,dx}$.Substitute $t = {\cos ^2}x$.Then,differentiating both sides with respect to $x$,we get $dt = 2\cos x(-\sin x)\,dx = -\sin 2x\,dx$.This implies $\sin 2x\,dx = -dt$.Substituting these into the integral,we get $I = \int {e^t(-dt)} = -\int {e^t\,dt}$.Integrating $e^t$,we get $I = -{e^t} + c$.Substituting back $t = {\cos ^2}x$,we get $I = -{e^{{{\cos }^2}x}} + c$.

$\int {{e^{{{\cos }^2}x}}\sin 2x\,dx = }$

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