Find the value of k if int cos ^k(x) sin (x) | vedclass

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(B) Given the integral: $\int \cos ^k(x) \sin (x) d x = \frac{-1}{4} \cos ^4(x) + C$.Let $I = \int \cos ^k(x) \sin (x) d x$.Substitute $t = \cos x$,th

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$3$

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(B) Given the integral: $\int \cos ^k(x) \sin (x) d x = \frac{-1}{4} \cos ^4(x) + C$.Let $I = \int \cos ^k(x) \sin (x) d x$.Substitute $t = \cos x$,then $dt = -\sin x d x$,which implies $\sin x d x = -dt$.Substituting these into the integral,we get:$I = \int t^k (-dt) = -\int t^k dt$.Integrating $t^k$ with respect to $t$ gives:$I = -\frac{t^{k+1}}{k+1} + C$.Substituting back $t = \cos x$:$I = -\frac{\cos^{k+1} x}{k+1} + C$.Comparing this with the given expression $\frac{-1}{4} \cos^4(x) + C$,we have:$\frac{1}{k+1} = \frac{1}{4}$ and $k+1 = 4$.Therefore,$k = 3$.

Find the value of $k$ if $\int \cos ^k(x) \sin (x) d x = \frac{-1}{4} \cos ^4(x) + C$.

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