If int e^{sin ^2 x}(sin x cos x+cos ^3 x sin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $I = \int e^{\sin ^2 x}(\sin x \cos x + \cos ^3 x \sin x) d x$. Factor out $\sin x \cos x$: $I = \int e^{\sin ^2 x}(1 + \cos ^2 x) \sin x \cos

Vedclass · Accepted Answer

$-\frac{1}{2} \sin 2 x$

Vedclass · Answer

(D) Let $I = \int e^{\sin ^2 x}(\sin x \cos x + \cos ^3 x \sin x) d x$. Factor out $\sin x \cos x$: $I = \int e^{\sin ^2 x}(1 + \cos ^2 x) \sin x \cos x d x$. Since $\cos ^2 x = 1 - \sin ^2 x$, we have: $I = \int e^{\sin ^2 x}(1 + 1 - \sin ^2 x) \sin x \cos x d x = \int e^{\sin ^2 x}(2 - \sin ^2 x) \sin x \cos x d x$. Let $t = \sin ^2 x$, then $dt = 2 \sin x \cos x d x$, so $\sin x \cos x d x = \frac{dt}{2}$. Substituting into the integral: $I = \frac{1}{2} \int e^t(2 - t) d t = \int e^t d t - \frac{1}{2} \int t e^t d t$. Using integration by parts for $\int t e^t d t = t e^t - e^t$: $I = e^t - \frac{1}{2}(t e^t - e^t) + C = e^t(1 - \frac{t}{2} + \frac{1}{2}) + C = e^t(\frac{3}{2} - \frac{t}{2}) + C$. Comparing with $e^{\sin ^2 x}(1 + f(x)) + C$, we have $1 + f(x) = \frac{3}{2} - \frac{\sin ^2 x}{2}$. Thus, $f(x) = \frac{1}{2} - \frac{\sin ^2 x}{2}$. Taking the derivative: $f^{\prime}(x) = 0 - \frac{1}{2}(2 \sin x \cos x) = -\sin x \cos x = -\frac{1}{2} \sin 2 x$.

If $\int e^{\sin ^2 x}(\sin x \cos x+\cos ^3 x \sin x) d x = e^{\sin ^2 x}(1+f(x))+c$, then $f^{\prime}(x)=$

Similar Questions

If $\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x=A \sin 2 x+B$,then $A$ is equal to

$\int \frac{x - 1}{(x + 1) \sqrt{x(x^2 + x + 1)}} dx =$

If $\int \frac{e^{\frac{x}{2}}}{\sqrt{e^{-x}-e^x}} \, dx = \sin^{-1}(f(x)) + C$,(where $C$ is the constant of integration),then $f(2)$ has the value:

If $f(x) = \int \left[ \tan^2 x + \cot^2 x + \frac{4(\sin^3 x + \cos^3 x)}{\sin^2 2x} \right] dx$ and $f\left(\frac{\pi}{4}\right) = 0$, then $3 \left[ f\left(\frac{\pi}{6}\right) + 2 \right] = $

Evaluate the integral: $\int \sqrt{x^2+4x+1} \, dx = \text{ . . . . . . } + C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module