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How do you integrate $\int{{{\sin }^{2}}x}$by integration by parts method?

seo-qna
Last updated date: 18th Jul 2024
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Answer
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Hint: In integration by parts method if we integrate $f\left( x \right)g\left( x \right)$ we can write
$\int{f\left( x \right)g\left( x \right)dx=f\left( x \right)\int{g\left( x \right)dx-\int{f'\left( x \right)}}\left( \int{g\left( x \right)dx} \right)dx}$ where$f'\left( x \right)=\dfrac{df\left( x \right)}{dx}$.
While choosing $f\left( x \right)$ and $g\left( x \right)$ choose in such a way that solving $\int{f'\left( x \right)}\left( \int{g\left( x \right)dx} \right)dx$ would be easier. For example while integrating $x{{e}^{x}}$ our $f\left( x \right)$ would be x and $g\left( x \right)$ will be ${{e}^{x}}$so that solving $\int{f'\left( x \right)}\left( \int{g\left( x \right)dx} \right)dx$ is easier.

Complete step by step answer:
The formula for integration by parts is $\int{f\left( x \right)g\left( x \right)dx=f\left( x \right)\int{g\left( x \right)dx-\int{f'\left( x \right)}}\left( \int{g\left( x \right)dx} \right)dx}$
Where$f'\left( x \right)=\dfrac{df\left( x \right)}{dx}$.
In the given question we have to integrate ${{\sin }^{2}}x$ by integration by parts method.
So $f\left( x \right)g\left( x \right)$=${{\sin }^{2}}x$
We can choose $f\left( x \right)=\sin x$ and $g\left( x \right)=\sin x$ so we can write
$\int{{{\sin }^{2}}xdx=\int{\sin x}}\times \sin xdx$
Substituting $f\left( x \right)$ as $\sin x$,$g\left( x \right)$ as $\sin x$ and $f\left( x \right)g\left( x \right)$ as ${{\sin }^{2}}x$
In the formula for integration by parts we get
$\int{{{\sin }^{2}}xdx=\sin x\int{\sin xdx-\int{\dfrac{d\sin x}{dx}\left( \int{\sin xdx} \right)}}}dx$
We know that $\dfrac{d\sin x}{dx}=\cos x$ and $\int{\sin xdx=-\cos x}$ substituting these values in the above equation
$\Rightarrow \int{{{\sin }^{2}}xdx=\sin x\left( -\cos x \right)-\int{\cos x\times \left( -\cos x \right)}}dx$
$\Rightarrow \int{{{\sin }^{2}}xdx=-\sin x\cos x+\int{{{\cos }^{2}}x}}dx$
We know that ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ substituting this value in the equation we get
$\Rightarrow \int{{{\sin }^{2}}x=-\sin x\cos x+\int{\left( 1-{{\sin }^{2}}x \right)dx}}$
Let’s assume $I=\int{{{\sin }^{2}}xdx}$
Substituting I in the equation
$I=-\sin x\cos x+\int{1dx-I}$
By solving the equation
$2I=-\sin x\cos x+\int{1dx}$
We know that
$\int{1dx=x+{{c}_{1}}}$
Where c1 is a constant so substituting the value in the equation we get
$2I=-\sin x\cos x+x+{{c}_{1}}$
Now we can write
$I=\dfrac{x-\cos x\sin x}{2}+c$
where c is a constant
$c=\dfrac{{{c}_{1}}}{2}$
As the final answer we can write
$\int{{{\sin }^{2}}x=\dfrac{x-\sin x\cos x}{2}+c}$
We can verify the result by solving the integration we can write
${{\sin }^{2}}x=\dfrac{1-\cos 2x}{2}$
And solve the integration very easily and it will match with our result.

Note: Always remember the formula for integrating in integration by parts method. Some people make mistakes while choosing f(x) and g(x) so carefully choose f(x) and g(x) such that integration would be easier to solve. Sometimes when we solve a problem we find the function of L.H.S in the right hand side in that case we should take our L.H.S as variable I like we did in the above question then it would be easier to solve we just have to find the value of I.