integration of sin7x

The integration of $sin^7x$, can be found using integration substitution and trigonometry identities . The integral of $\sin^7 x $ with respect to (x) is:

\[
\int \sin^7 x \, dx =\frac{\cos^7(x)}{7} – \frac{3\cos^5(x)}{5} + \cos^3(x) – \cos(x) + C
\]

Here, (C) represents the constant of integration, which is added because the process of integration determines the antiderivative up to an arbitrary constant.

Proof of integration of sin^7x

The integration can be written as

\[
\int \sin^7 x \, dx = \int \sin^6 x \sin x \, dx
\]

Now $\sin^2 x = 1 – cos^2 x$, so this can be written as

$=\int (1-\cos^2 x)^3 \sin x \, dx

Now using integration by substitution

$\cosx = t$
$ -\sin x dx =dt$
So, this can be written as

$= -\int (1-t^2)^3 \, dt$

Now $(a-b)^3 = a^3 -3a^2 b + 3ab^2 -b^3$

$=\int [1 – t^6 -3t^2 + 3t^4] \, dt$
$ =- t + \frac {t^7}{7} + t^3 – \frac {3t^5}{5} + C$

Substituting back

\[
\int \sin^7 x \, dx =\frac{\cos^7(x)}{7} – \frac{3\cos^5(x)}{5} + \cos^3(x) – \cos(x) + C
\]

Proof of integration of sin(7x)

we need to find the below integral

\[
\int \sin(7x) \, dx
\]

let $7x = t$
$7 dx=dt$
$dx= \frac {dt}{7}$
Therefore

\[
= \frac {1}{7} \int \sin(t) \, dt = -\frac {\cos t}{7} + C
\]

Substituting back

\[
= -\frac {\cos 7x}{7} + C
\]

Solved Examples

Question 1

$\int \frac {\cos x}{\sin^7 x} \; dx $

Solution

$\sinx = t$
$ \cos x dx =dt$
So, this can be written as

$= \int \frac {1}{t^7} \; dt$
$= -\frac {1}{6t^6} + C$

Substituting back

$=- \frac {1}{6 \sin^6 x} + C$