Question

Obtain the reduction formula for $\int {{{\sin }^n}x\,dx}$ an integer $n \ge 2$ and deduce the value of$\int {{{\sin }^4}x\,dx}$.

Answer 1

Hint : Split${\sin ^n}x$ in to two parts and apply the integration of the $uv{\rm{ }}dx$ formula and deduce the value of the $\int {{{\sin }^4}x\,dx}$. Sin, cos, tan, cosec, cot, and sec are the trigonometric values that are used to find the angles or radians in any concepts, they are further differentiated and integrated. If we differentiate any value, then the value will decrease and it may result in the zero but if we integrate any value then the degree will be increased and the value will be increased.

Complete step-by-step answer :
Given:
The trigonometric expression is ${I_n} = \int {{{\sin }^n}x\,dx}$.
The value of integer is $n \ge 2$ .
To find:
The value of $\int {{{\sin }^4}x\,dx}$.
Reduce the equation by splitting the above equation ${I_n} = \int {{{\sin }^n}x\,dx}$.
${{\rm{I}}_{\rm{n}}}{\rm{ = }}\int {{\rm{sinx}} \cdot {\rm{si}}{{\rm{n}}^{{\rm{n - 1}}}}{\rm{x}}\,{\rm{dx}}}$
Now, we will apply the equation $\int {uv\,dx = \int {u\,dv + \int {v\,d} } } u$, then

{I_n} = - \cos ({\sin ^{n - 1}}x) - \int { - \cos x(n - 1).{{\sin }^{n - 2}}x\,\cos x\,dx} \\\ {I_n} = - {\sin ^{n - 1}}x\,\cos x + (n - 1)\int {{{\sin }^{n - 2}}x(1 - {{\sin }^2}x)\,dx} \\\ {I_n} = - {\sin ^{n - 1}}x\,\cos x + (n - 1)\int {{{\sin }^{n - 2}}xdx - (n - 1)\int {{{\sin }^n}x} \,dx} \\\ {I_n} = - {\sin ^{n - 1}}x\,\cos x + (n - 1)\int {{{\sin }^{n - 2}}xdx - (n - 1){I_n}} \\\ {I_n} + (n - 1){I_n} = - {\sin ^{n - 1}}x\,\cos x + (n - 1)\int {{{\sin }^{n - 2}}} xdx \end{array}$$ Now, we know that the formula for $${I_n} = \int {{{\sin }^n}x\,dx} $$ is $$\dfrac{{ - 1}}{n}{\sin ^{n - 1}}x\,\cos x + \dfrac{{(n - 1)}}{n}\int {{{\sin }^{n - 2}}} xdx$$. Now, by deducing the equation $$\int {{{\sin }^4}x\,dx} $$, we get, $$\begin{array}{c} \int {{{\sin }^4}x\,dx} = \dfrac{{ - 1}}{4}{\sin ^3}x\,\cos x + \dfrac{3}{4}\int {{{\sin }^2}} xdx\\\ = \dfrac{{ - 1}}{4}{\sin ^3}x\,\cos x + \dfrac{3}{4}\int {\dfrac{{1 - \cos 2x}}{2}} dx\\\ = \dfrac{{ - 1}}{4}{\sin ^3}x\,\cos x + \dfrac{3}{8}\int {dx} - \dfrac{3}{8}\int {\cos 2xdx} \\\ = \dfrac{{ - 1}}{4}{\sin ^3}x\,\cos x + \dfrac{{3x}}{8} - \dfrac{{3(\sin 2x)}}{16} + c \end{array}$$ Therefore, the deduction of $$\int {{{\sin }^4}x\,dx} $$ is $$\dfrac{{ - 1}}{4}{\sin ^3}x\,\cos x + \dfrac{{3x}}{8} - \dfrac{{3(\sin 2x)}}{16} + c$$. **Note** : Make sure while solving every step, because the solution involves some complex derivations and we also need to be aware about the process of solving. One has to deduce the formula for the $$\int {{{\sin }^n}x\,dx} $$first and then substitute the values of the $$\int {{{\sin }^4}x\,dx} $$ in the reduced formula. We have to be sure about the integration formulas like if we integrate the $ \sin x $ then, the result will be $ \cos x $ , so there will be many formulas that help in the integration to solve the many problems, so be aware about each and every formula in the integration.