Question

Integration of $si{n}^{7}xdx$.

Answer 1

Step 1: Separate the expression

The given expression is$si{n}^{7}xdx$.

Let its integration be$I$.

$\therefore$ $I=\int si{n}^{7}xdx$

$=\int si{n}^{6}x.\sin xdx$

$=\int {\left({\sin }^{2}x\right)}^3.\sin xdx$

Step 2: Substitute the trigonometric and algebraic identities

We know that ${\sin }^{2}x=1-{\cos }^{2}x$.

$\therefore$$I=\int {\left(1-{\cos }^{2}x\right)}^3.\sin xdx$

We know that${\left(a-b\right)}^3=a^3-3a^{2}b+3a{b}^2-b^3$.

$\therefore$$I=\int \left(1-{\cos }^{6}x-3{\cos }^{2}x+3{\cos }^{4}x\right).\sin xdx$

$=\int \sin xdx-\int {\cos }^{6}x.\sin xdx-3\int {\cos }^{2}x.\sin xdx+3\int {\cos }^{4}x.\sin xdx$

Step 3: Substitute a variable and solve

Let $y=\cos x$

$\Rightarrow$$dy=-\sin xdx$

$\therefore$ $I=-\cos x+\int y^{6}dy+3\int y^{2}dy-3\int y^{4}dy$

$=-\cos x+\frac{y^7}{7}+3\frac{y^3}{3}-3\frac{y^5}{5}+c$ where $c$ is the constant of integration.

$=-\cos x+\frac{{\cos }^{7}x}{7}+{\cos }^{3}x-\frac{3}{5}{\cos }^{5}x+c$

Hence, when $si{n}^{7}xdx$ is integrated we get $-\cos x+\frac{{\cos }^{7}x}{7}+{\cos }^{3}x-\frac{3}{5}{\cos }^{5}x+c$.