What is the integration of sin^7 dx?

1.	What is the integration of sin^7 dx?
Answer» 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 $\displaystyle\mathrm{If}~\frac{dF\left(x\right)}{dx}=f\left(x\right)\mathrm{\:then\:}\int f\left(x\right)dx=F\left(x\right)+C$ $\displaystyle I=-\cos \left(x\right)+\cos ^3\left(x\right)-\frac{3\cos ^5\left(x\right)}{5}+\frac{\cos ^7\left(x\right)}{7}+C$

Answer»

Answer:

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