Mathematics

# Evaluate $\int\limits_0^{\frac{\pi }{4}} {{{\cos }^{\frac{3}{2}}}\left( {2x} \right)cos\left( x \right)dx}$

##### SOLUTION

Consider the given integral.

$I=\int_{0}^{\frac{\pi }{4}}{{{\cos }^{3/2}}2x\cos x}dx$

$I=\int_{0}^{\frac{\pi }{4}}{{{\left( \cos 2x \right)}^{3/2}}\cos x}dx$

$I=\int_{0}^{\frac{\pi }{4}}{{{\left( 1-2{{\sin }^{2}}x \right)}^{3/2}}\cos x}dx$

Let $t=\sin x$

$dt=\cos xdx$

Therefore,

$I=\int_{0}^{\frac{1}{\sqrt{2}}}{{{\left( 1-2{{t}^{2}} \right)}^{3/2}}dt}$

$I=\int_{0}^{\frac{1}{\sqrt{2}}}{{{\left( 1-{{\left( \sqrt{2}t \right)}^{2}} \right)}^{3/2}}dt}$

Let $\sqrt{2}t=\sin u$

$\sqrt{2}\dfrac{dt}{du}=\cos u$

$dt=\dfrac{\cos udu}{\sqrt{2}}$

Therefore,

$I=\dfrac{1}{\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{{{\left( 1-{{\sin }^{2}}u \right)}^{3/2}}\cos udu}$

$I=\dfrac{1}{\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{{{\left( {{\cos }^{2}}u \right)}^{3/2}}\cos udu}$

$I=\dfrac{1}{\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{{{\cos }^{4}}udu}$

$I=\dfrac{1}{\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{{{\left( {{\cos }^{2}}u \right)}^{2}}du}$

$I=\dfrac{1}{\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{{{\left( \dfrac{1+\cos 2u}{2} \right)}^{2}}du}$

$I=\dfrac{1}{4\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{\left( 1+{{\cos }^{2}}2u+2\cos 2u \right)du}$

$I=\dfrac{1}{4\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{\left( 1+\left( \dfrac{1+\cos 4u}{2} \right)+2\cos 2u \right)du}$

$I=\dfrac{1}{4\sqrt{2}}\int_{0}^{\frac{\pi }{2}}{\left( 1+\dfrac{1}{2}+\dfrac{\cos 4u}{2}+2\cos 2u \right)du}$

$I=\dfrac{1}{4\sqrt{2}}\left[ u+\dfrac{u}{2}+\dfrac{1}{2}\left( \dfrac{\sin 4u}{4} \right)+2\dfrac{\sin 2u}{2} \right]_{0}^{\frac{\pi }{2}}$

$I=\dfrac{1}{4\sqrt{2}}\left[ \dfrac{3u}{2}+\dfrac{\sin 4u}{8}+\sin 2u \right]_{0}^{\frac{\pi }{2}}$

$I=\dfrac{1}{4\sqrt{2}}\left[ \dfrac{3\pi }{4}+\dfrac{\sin 4\left( \dfrac{\pi }{2} \right)}{8}+\sin 2\left( \dfrac{\pi }{2} \right)-\left( 0 \right) \right]$

$I=\dfrac{1}{4\sqrt{2}}\left[ \dfrac{3\pi }{4}+\dfrac{\sin \left( 2\pi \right)}{8}+\sin \left( \pi \right) \right]$

$I=\dfrac{1}{4\sqrt{2}}\left[ \dfrac{3\pi }{4}+0+0 \right]$

$I=\dfrac{3\pi }{16\sqrt{2}}$

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Subjective Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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