int sin^5 x cos^4 x , dx = | vedclass

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

Question

(A) Let $I = \int \sin^5 x \cos^4 x \, dx$. Substitute $\cos x = t$,then $-\sin x \, dx = dt$,or $\sin x \, dx = -dt$. We can write $\sin^5 x$ as $(\s

Vedclass · Accepted Answer

$-\frac{1}{5}\cos^5 x + \frac{2}{7}\cos^7 x - \frac{1}{9}\cos^9 x + c$

Vedclass · Answer

(A) Let $I = \int \sin^5 x \cos^4 x \, dx$. Substitute $\cos x = t$,then $-\sin x \, dx = dt$,or $\sin x \, dx = -dt$. We can write $\sin^5 x$ as $(\sin^2 x)^2 \sin x = (1 - \cos^2 x)^2 \sin x$. Substituting these into the integral: $I = \int (1 - t^2)^2 t^4 (-dt) = - \int (1 - 2t^2 + t^4) t^4 \, dt$ $I = - \int (t^4 - 2t^6 + t^8) \, dt$ $I = - (\frac{t^5}{5} - \frac{2t^7}{7} + \frac{t^9}{9}) + c$ $I = - \frac{t^5}{5} + \frac{2t^7}{7} - \frac{t^9}{9} + c$ Substituting $t = \cos x$ back: $I = - \frac{1}{5}\cos^5 x + \frac{2}{7}\cos^7 x - \frac{1}{9}\cos^9 x + c$.

$\int \sin^5 x \cos^4 x \, dx = $

Similar Questions

For $x < 1$,evaluate $\int \frac{x-x^2}{\sqrt{1-x}} d x$.

$\int {\sqrt {{e^x} - 1} } dx = $

$\int \left( \frac{(\sin^4 x + 2 \cos^2 x - 1) \cos x}{(1 + \sin x)^6} \right) dx =$

If $\int \frac{\sin x}{3+4 \cos ^2 x} \,dx = A \tan ^{-1}(B \cos x) + C$, (where $C$ is a constant of integration), then the value of $A+B$ is

$\int \sin \sqrt{x} \,d x=\ldots+C$ (where $C$ is a constant of integration.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module