int cos ^3 x cdot e^{log (sin x)} d x= | vedclass

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

Question

(A) Given the integral $I = \int \cos ^3 x \cdot e^{\log (\sin x)} d x$. Since $e^{\log (f(x))} = f(x)$,the integral simplifies to: $I = \int \cos ^3

Vedclass · Accepted Answer

$\frac{-\cos ^4 x}{4}+c$

Vedclass · Answer

(A) Given the integral $I = \int \cos ^3 x \cdot e^{\log (\sin x)} d x$. Since $e^{\log (f(x))} = f(x)$,the integral simplifies to: $I = \int \cos ^3 x \cdot \sin x d x$. Let $u = \cos x$. Then $du = -\sin x d x$,which implies $\sin x d x = -du$. Substituting these into the integral: $I = \int u^3 (-du) = -\int u^3 du$. Integrating $u^3$ gives $\frac{u^4}{4}$. Thus,$I = -\frac{u^4}{4} + c$. Substituting back $u = \cos x$,we get: $I = -\frac{\cos ^4 x}{4} + c$.

$\int \cos ^3 x \cdot e^{\log (\sin x)} d x=$

Similar Questions

$\int \frac{e^{x}(1+x)}{\cos ^{2}(x e^{x})} d x$ equals

$\int \frac{\csc x}{\log \tan \frac{x}{2}} \, dx = $

$\int \tan^3 2x \sec 2x \, dx = $

For ${x^2} \ne n\pi + 1, n \in N$ (the set of natural numbers),the integral $\int {x\sqrt {\frac{{2\sin \left( {{x^2} - 1} \right) - \sin 2\left( {{x^2} - 1} \right)}}{{2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} } dx$ is

$\int 7^{7^{7^{x}}} 7^{7^{x}} 7^{x} \,d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module