# Integrate the following function:

\[\sqrt{\sin 2x}\cos 2x\]

Answer

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Hint: First of all, take sin 2x = t. By differentiating it, we get \[cos2x.2=dt\]. Now put all the terms of ‘x’ that are sin 2x and cos 2x dx in terms of ‘t’ in the given integral. Then integrate using the formula \[\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}\].

Here, we have to integrate the function \[\sqrt{\sin 2x}\cos 2x\].

Let us take the integral given in the question as,

\[I=\int{\sqrt{\sin 2x}\cos 2x}\text{ }dx\]

Let us consider, \[\sin 2x=t.....\left( i \right)\]

We know that, \[\dfrac{d}{dx}\left( \sin x \right)=\cos x\]

Also, by chain rule, if \[y=f\left( u \right)\] and \[u=g\left( x \right)\], then \[\dfrac{dy}{dx}={{f}^{'}}\left( u \right).\dfrac{du}{dx}\]

Therefore, by differentiating equation (i) on both sides with respect to t, we get,

\[\cos 2x.\dfrac{d}{dt}\left( 2x \right)=1\]

By further solving the above expression, we get

\[\left( \cos 2x \right).2.\dfrac{dx}{dt}=1\]

By multiplying dt on both sides, we get,

\[\left( \cos 2x \right).2.dx=dt\]

By dividing 2 on both sides, we get,

\[\left( \cos 2x \right).dx=\dfrac{dt}{2}\]

Now, we will put the value of sin 2x and cos 2x dx in terms of ‘t’ in the given integral, we will get

\[I=\int{\sqrt{t}}\dfrac{dt}{2}\]

Or, we can write the above expression as,

\[I=\dfrac{1}{2}\int{{{t}^{\dfrac{1}{2}}}}dt\]

Now, we know that \[\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}\]. By applying this in the above integral, we get,

\[I=\dfrac{1}{2}\dfrac{\left( {{t}^{\dfrac{1}{2}+1}} \right)}{\left( \dfrac{1}{2}+1 \right)}+k\]

By simplifying the above integral, we get

\[I=\left( \dfrac{1}{2} \right)\dfrac{{{t}^{\dfrac{3}{2}}}}{\left( \dfrac{3}{2} \right)}+k\]

By cancelling the like terms, we get,

\[I=\dfrac{{{t}^{\dfrac{3}{2}}}}{3}+k\]

Now, we will replace it with sin 2x as we have assumed earlier. Therefore, we will get,

\[I=\dfrac{{{\left( \sin 2x \right)}^{\dfrac{3}{2}}}}{3}+k\]

Therefore, our required integration of function \[\sqrt{\sin 2x}\cos 2x\] is \[\dfrac{{{\left( \sin 2x \right)}^{\dfrac{3}{2}}}}{3}+k\].

Note: Students should always remember to convert the assumed variable back to the original variable like here, students must convert ‘t’ back to ‘x’ at the end of the solution. Also, students can cross check their answer by differentiating the final answer and checking if it is giving the expression given in the question or not.

Here, we have to integrate the function \[\sqrt{\sin 2x}\cos 2x\].

Let us take the integral given in the question as,

\[I=\int{\sqrt{\sin 2x}\cos 2x}\text{ }dx\]

Let us consider, \[\sin 2x=t.....\left( i \right)\]

We know that, \[\dfrac{d}{dx}\left( \sin x \right)=\cos x\]

Also, by chain rule, if \[y=f\left( u \right)\] and \[u=g\left( x \right)\], then \[\dfrac{dy}{dx}={{f}^{'}}\left( u \right).\dfrac{du}{dx}\]

Therefore, by differentiating equation (i) on both sides with respect to t, we get,

\[\cos 2x.\dfrac{d}{dt}\left( 2x \right)=1\]

By further solving the above expression, we get

\[\left( \cos 2x \right).2.\dfrac{dx}{dt}=1\]

By multiplying dt on both sides, we get,

\[\left( \cos 2x \right).2.dx=dt\]

By dividing 2 on both sides, we get,

\[\left( \cos 2x \right).dx=\dfrac{dt}{2}\]

Now, we will put the value of sin 2x and cos 2x dx in terms of ‘t’ in the given integral, we will get

\[I=\int{\sqrt{t}}\dfrac{dt}{2}\]

Or, we can write the above expression as,

\[I=\dfrac{1}{2}\int{{{t}^{\dfrac{1}{2}}}}dt\]

Now, we know that \[\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}\]. By applying this in the above integral, we get,

\[I=\dfrac{1}{2}\dfrac{\left( {{t}^{\dfrac{1}{2}+1}} \right)}{\left( \dfrac{1}{2}+1 \right)}+k\]

By simplifying the above integral, we get

\[I=\left( \dfrac{1}{2} \right)\dfrac{{{t}^{\dfrac{3}{2}}}}{\left( \dfrac{3}{2} \right)}+k\]

By cancelling the like terms, we get,

\[I=\dfrac{{{t}^{\dfrac{3}{2}}}}{3}+k\]

Now, we will replace it with sin 2x as we have assumed earlier. Therefore, we will get,

\[I=\dfrac{{{\left( \sin 2x \right)}^{\dfrac{3}{2}}}}{3}+k\]

Therefore, our required integration of function \[\sqrt{\sin 2x}\cos 2x\] is \[\dfrac{{{\left( \sin 2x \right)}^{\dfrac{3}{2}}}}{3}+k\].

Note: Students should always remember to convert the assumed variable back to the original variable like here, students must convert ‘t’ back to ‘x’ at the end of the solution. Also, students can cross check their answer by differentiating the final answer and checking if it is giving the expression given in the question or not.

Last updated date: 01st Oct 2023

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