Questions & Answers

Question

Answers

Answer
Verified

Hint: In order to solve this integration we have to need some trigonometry relationship from which we can solve it easily and more accurate way we will use some relationship like $\left( {\dfrac{{1 - cos2x}}{2}} \right) = {\sin ^2}x$ to solve this problem.

Complete step-by-step answer:

Given that $\int {\sin x\sqrt {1 - \cos 2x} } dx$

Let $\left( {\dfrac{{1 - cos2x}}{2}} \right) = {\sin ^2}x$$I = \int {\sin x\sqrt {1 - \cos 2x} } dx \to (1)$

$\because \left( {\dfrac{{1 - cos2x}}{2}} \right) = {\sin ^2}x$

By cross multiplying we get

$1 - \cos 2x = 2{\sin ^2}x$

We can put this value in equation $(1)$ in the place of $1 - \cos 2x$

So we get

$I = \int {\sin x\sqrt {2{{\sin }^2}x} } dx$

Here we can take $\sqrt 2 $common so equation becomes

$I = \sqrt 2 \int {\sin x\sqrt {{{\sin }^2}x} } dx$

We can write $\sqrt {{{\sin }^2}x} $ as $\sin x$

So further solving it

$

I = \sqrt 2 \int {\sin x(\sin x)} dx \\

= \sqrt 2 \int {{{\sin }^2}x} dx \\

$

Again we can write ${\sin ^2}x$ as $\left( {\dfrac{{1 - cos2x}}{2}} \right)$

$I = \sqrt 2 \int {\left( {\dfrac{{1 - cos2x}}{2}} \right)} dx$

Further solving it

$I = \dfrac{1}{{\sqrt 2 }}\int {\left( {1 - \cos 2x} \right)} dx$

By splitting it we get

$I = \dfrac{1}{{\sqrt 2 }}\int {dx - \dfrac{1}{{\sqrt 2 }}\int {\cos 2xdx} } $

$\because \int {f'(ax + b)dx = \dfrac{{f(ax + b)}}{a} + c} $

$\because \int {dx = x{\text{ and }}\int {\cos 2xdx = } \dfrac{{\sin 2x}}{2}} + c$

By replacing it we get

$I = \dfrac{1}{{\sqrt 2 }}(x) - \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{{\sin 2x}}{2}} \right) + c$

Where $c$ is the constant.

So the final term we get by solving $\int {\sin x\sqrt {1 - \cos 2x} } dx$ is $I = \dfrac{1}{{\sqrt 2 }}(x) - \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{{\sin 2x}}{2}} \right) + c$

Note- Integration is a way of adding slices to find the whole. Integration can be used to find areas, volumes, central points and many useful things. But it is easiest to start with finding the area under the curve of a function.

Complete step-by-step answer:

Given that $\int {\sin x\sqrt {1 - \cos 2x} } dx$

Let $\left( {\dfrac{{1 - cos2x}}{2}} \right) = {\sin ^2}x$$I = \int {\sin x\sqrt {1 - \cos 2x} } dx \to (1)$

$\because \left( {\dfrac{{1 - cos2x}}{2}} \right) = {\sin ^2}x$

By cross multiplying we get

$1 - \cos 2x = 2{\sin ^2}x$

We can put this value in equation $(1)$ in the place of $1 - \cos 2x$

So we get

$I = \int {\sin x\sqrt {2{{\sin }^2}x} } dx$

Here we can take $\sqrt 2 $common so equation becomes

$I = \sqrt 2 \int {\sin x\sqrt {{{\sin }^2}x} } dx$

We can write $\sqrt {{{\sin }^2}x} $ as $\sin x$

So further solving it

$

I = \sqrt 2 \int {\sin x(\sin x)} dx \\

= \sqrt 2 \int {{{\sin }^2}x} dx \\

$

Again we can write ${\sin ^2}x$ as $\left( {\dfrac{{1 - cos2x}}{2}} \right)$

$I = \sqrt 2 \int {\left( {\dfrac{{1 - cos2x}}{2}} \right)} dx$

Further solving it

$I = \dfrac{1}{{\sqrt 2 }}\int {\left( {1 - \cos 2x} \right)} dx$

By splitting it we get

$I = \dfrac{1}{{\sqrt 2 }}\int {dx - \dfrac{1}{{\sqrt 2 }}\int {\cos 2xdx} } $

$\because \int {f'(ax + b)dx = \dfrac{{f(ax + b)}}{a} + c} $

$\because \int {dx = x{\text{ and }}\int {\cos 2xdx = } \dfrac{{\sin 2x}}{2}} + c$

By replacing it we get

$I = \dfrac{1}{{\sqrt 2 }}(x) - \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{{\sin 2x}}{2}} \right) + c$

Where $c$ is the constant.

So the final term we get by solving $\int {\sin x\sqrt {1 - \cos 2x} } dx$ is $I = \dfrac{1}{{\sqrt 2 }}(x) - \dfrac{1}{{\sqrt 2 }}\left( {\dfrac{{\sin 2x}}{2}} \right) + c$

Note- Integration is a way of adding slices to find the whole. Integration can be used to find areas, volumes, central points and many useful things. But it is easiest to start with finding the area under the curve of a function.

×

Sorry!, This page is not available for now to bookmark.