Question

What is the integral of ${{\sin }^{6}}x$?

Answer 1

From the given question we have been asked to find the integral of ${{\sin }^{6}}x$. For solving this question we will use the concept of integration and its formulae. We also use the formulae in trigonometry like $2{{\sin }^{2}}\theta =1-\cos 2\theta$ and etc.First we will take the given trigonometric expression and then we simplify the expression using trigonometry formulae and basic mathematical operations like addition and subtraction and then we will apply the integral and solve the given equation.

Complete step-by-step solution:
We are asked to find $\int{{{\sin }^{6}}x}$
But firstly we take the expression and simplify it.
$\Rightarrow {{\sin }^{6}}x={{\left( {{\sin }^{2}}x \right)}^{2}}{{\sin }^{2}}x$
Now we will use the basic trigonometric formulae which is $2{{\sin }^{2}}\theta =1-\cos 2\theta$ and substitute in the above equation. so, we get the equation reduced as follows.
$\Rightarrow {{\sin }^{6}}x={{\left( \dfrac{1-\cos 2x}{2} \right)}^{2}}\left( \dfrac{1-\cos 2x}{2} \right)$
Now we will expand the square.
$\Rightarrow {{\sin }^{6}}x=\dfrac{1}{8}\left( 1-2\cos 2x+{{\cos }^{2}}2x \right)\left( 1-\cos 2x \right)$
Now we will use the basic trigonometric formulae which is ${{\cos }^{2}}x=1+\cos 2x$ and substitute in the above equation. so, we get the equation reduced as follows.
$\Rightarrow {{\sin }^{6}}x=\dfrac{1}{8}\left( 1-2\cos 2x+\dfrac{1+\cos 4x}{2} \right)\left( 1-\cos 2x \right)$
$\Rightarrow {{\sin }^{6}}x=\dfrac{1}{16}\left( 2-4\cos 2x+1+\cos 4x \right)\left( 1-\cos 2x \right)$
$\Rightarrow {{\sin }^{6}}x=\dfrac{1}{16}\left( 3-4\cos 2x+\cos 4x \right)\left( 1-\cos 2x \right)$
Now we will use basic multiplication operations and we will expand the brackets.
$\Rightarrow {{\sin }^{6}}x=\dfrac{1}{16}\left( 3-7\cos 2x+\cos 4x+2\times 2{{\cos }^{2}}2x-\dfrac{1}{2}2\cos 4x\cos 2x \right)$
Now we will use the basic trigonometric formulae which is $2\cos A\cos B=\cos \left( A+B \right)\cos \left( A-B \right)$ and substitute in the above equation. so, we get the equation reduced as follows.
$\Rightarrow {{\sin }^{6}}x=\dfrac{1}{16}\left( 3-7\cos 2x+\cos 4x+2\left( 1+\cos 4x \right)-\dfrac{1}{2}\left( cos6x+\cos 2x \right) \right)$
$\Rightarrow {{\sin }^{6}}x=\dfrac{1}{16}\left( 3-7\cos 2x+\cos 4x+\left( 2+2\cos 4x \right)-\dfrac{1}{2}\left( cos6x+\cos 2x \right) \right)$
$\Rightarrow {{\sin }^{6}}x=\dfrac{1}{32}\left( 6-14\cos 2x+2\cos 4x+\left( 4+4\cos 4x \right)-\left( cos6x+\cos 2x \right) \right)$
$\Rightarrow {{\sin }^{6}}x=\dfrac{1}{32}\left( 10-15\cos 2x+6\cos 4x-\cos 6x \right)$
Now the given expression is reduced to this equation. Now, we will apply integral to this expression. So, we get the integral as follows.
$\Rightarrow I=\dfrac{1}{32}\int{\left( 10-15\cos 2x+6\cos 4x-\cos 6x \right)}dx$
We know that $\int{\cos xdx=\sin x}$ and $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$.
$\Rightarrow I=\dfrac{1}{32}\left( 10x-\dfrac{15\sin 2x}{2}+\dfrac{6\sin 4x}{4}-\dfrac{\sin 6x}{6} \right)+c$
$\Rightarrow I=\dfrac{1}{32}\left( 10x-\dfrac{15\sin 2x}{2}+\dfrac{3\sin 4x}{2}-\dfrac{\sin 6x}{6} \right)+c$
$\Rightarrow I=\dfrac{1}{192}\left( 60x-45\sin 2x+9\sin 4x-\sin 6x \right)+c$

Note: Students must be very careful in doing the calculations. Students must have good knowledge in the concept of integration and trigonometry and their formulae. To solve these kind of questions we must know formulae like,

& \Rightarrow 2{{\sin }^{2}}\theta =1-\cos 2\theta \\\ & \Rightarrow {{\cos }^{2}}x=1+\cos 2x \\\ & \Rightarrow 2\cos A\cos B=\cos \left( A+B \right)\cos \left( A-B \right) \\\ & \Rightarrow \infty \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}} \\\ \end{aligned}$$