If I = int_0^{pi /4} sin^2 x , dx and J = int | vedclass

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(A) Given $I = \int_0^{\pi /4} \sin^2 x \, dx$ and $J = \int_0^{\pi /4} \cos^2 x \, dx$. Adding the two integrals: $I + J = \int_0^{\pi /4} (\sin^2 x

Vedclass · Accepted Answer

$\frac{\pi}{4} - J$

Vedclass · Answer

(A) Given $I = \int_0^{\pi /4} \sin^2 x \, dx$ and $J = \int_0^{\pi /4} \cos^2 x \, dx$. Adding the two integrals: $I + J = \int_0^{\pi /4} (\sin^2 x + \cos^2 x) \, dx$. Using the trigonometric identity $\sin^2 x + \cos^2 x = 1$,we get: $I + J = \int_0^{\pi /4} 1 \, dx$. Evaluating the integral: $I + J = [x]_0^{\pi /4} = \frac{\pi}{4} - 0 = \frac{\pi}{4}$. Therefore,$I = \frac{\pi}{4} - J$.

If $I = \int_0^{\pi /4} \sin^2 x \, dx$ and $J = \int_0^{\pi /4} \cos^2 x \, dx$,then $I = $

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