Question

what is the integration of sin^7 dx? ​

Answer 1

Answer. Hii machiii ✋. ➡️ sin^7 (x) dx = ∫ sin ^6 (x) sin(x) dx. Let u = cos(x), du = -sin( x) dx = ∫ {1 - cos²(x)}³ sin(x) dx.

Answer 2

Answer:-

$\displaystyle\int \sin ^7\left(x\right)dx=-\cos \left(x\right)+\cos ^3\left(x\right)-\frac{3\cos ^5\left(x\right)}{5}+\frac{\cos ^7\left(x\right)}{7}+C$

Step-by-step explanation:-

$\displaystyle I=\int \sin ^7\left(x\right)dx$

$\gray{\sin ^7\left(x\right)=\sin ^6\left(x\right)\sin \left(x\right)}$

$\displaystyle I=\int \sin ^6\left(x\right)\sin \left(x\right)dx$

$\gray{\sin ^6\left(x\right)=\left(\sin ^2\left(x\right)\right)^3}$

$\displaystyle I=\int \left(\sin ^2\left(x\right)\right)^3\sin \left(x\right)dx$

Use the following identity:    $\quad\sin ^2\left(x\right)=1-\cos ^2\left(x\right)$

$\displaystyle I=\int \left(1-\cos ^2\left(x\right)\right)^3\sin \left(x\right)dx$

Let’s substitute. Let $u=\cos \left(x\right)$

$\displaystyle I=\int \:-\left(1-u^2\right)^3du$

Expand: $-\left(1-u^2\right)^3$:

Apply Perfect Cube Formula:    $\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3$

$a=1,\:\:b=u^2$

$=1^3-3\cdot \:1^2u^2+3\cdot \:1\cdot \left(u^2\right)^2-\left(u^2\right)^3$

$\displaystyle \implies I = \int \:-1+3u^2-3u^4+u^6du$

Apply The SUm Rule of integration:

$\displaystyle\int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$\displaystyle I=-\int \:1du+\int \:3u^2du-\int \:3u^4du+\int \:u^6du$

$\displaystyle\implies I=-u+u^3-\frac{3u^5}{5}+\frac{u^7}{7}$

Substitute back $u=\cos \left(x\right)$

$\displaystyle I=-\cos \left(x\right)+\cos ^3\left(x\right)-\frac{3\cos ^5\left(x\right)}{5}+\frac{\cos ^7\left(x\right)}{7}$

Add a constant to the solution

$\displaystyle\mathrm{If}~\frac{dF\left(x\right)}{dx}=f\left(x\right)\mathrm{\:then\:}\int f\left(x\right)dx=F\left(x\right)+C$

$\displaystyle I=-\cos \left(x\right)+\cos ^3\left(x\right)-\frac{3\cos ^5\left(x\right)}{5}+\frac{\cos ^7\left(x\right)}{7}+C$