# Evaluate the given Integral:

$\int (\sqrt{\mathrm{cos}x})sinxdx$

Last updated date: 16th Mar 2023

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Answer

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Hint: Integral expression contains a function and its derivative, hence we use substitution method to reduce the integral into standard form.

Consider the expression,

$\int (\sqrt{\mathrm{cos}x})sinxdx$

It consists of two functions, where one function i.e. $\mathrm{cos}x$ is the derivative of another which is $\mathrm{sin}x$.

So, we can use the method of integration by substitution.

In this method we substitute one of the functions to reduce the expression into standard form.

Now let us consider, $\mathrm{cos}x=t$

Differentiating both sides with respect to x, we get

We know,$\frac{d(\mathrm{cos}x)}{dx}}=-\mathrm{sin}x$

Now,

$\frac{d(\mathrm{cos}x)}{dx}}\text{}={\displaystyle \frac{dt}{dx}$

$-\mathrm{sin}x={\displaystyle \frac{dt}{dx}}$

$-\mathrm{sin}xdx=dt$

Substitute $\mathrm{sin}xdx=-dt$ in the expression, we get

$\text{}\int (\sqrt{\mathrm{cos}x})sinxdx=-\int (\sqrt{t}\text{})dt$

We know, $\int {x}^{n}dx=\text{}{\displaystyle \frac{{x}^{n+1}}{n+1}}+C$

So, after integrating we get

$\text{}-\int (\sqrt{t}\text{})dt=-{\displaystyle \frac{{t}^{\frac{3}{2}}}{{\displaystyle \frac{3}{2}}}}+C$

$\text{}-\int (\sqrt{t}\text{})dt=-{\displaystyle \frac{2}{3}}\text{}{t}^{\frac{3}{2}}+C$

Re-substitute the value of t in terms of x, we get

$\text{}\int (\sqrt{\mathrm{cos}x})sinxdx=-{\displaystyle \frac{2}{3}}\text{}{\mathrm{cos}}^{\frac{3}{2}}x+C$

Note: Whenever an integrating expression consists of more than one function convert it into standard form by reduction method of integration such as substitution method of integration

Consider the expression,

$\int (\sqrt{\mathrm{cos}x})sinxdx$

It consists of two functions, where one function i.e. $\mathrm{cos}x$ is the derivative of another which is $\mathrm{sin}x$.

So, we can use the method of integration by substitution.

In this method we substitute one of the functions to reduce the expression into standard form.

Now let us consider, $\mathrm{cos}x=t$

Differentiating both sides with respect to x, we get

We know,$\frac{d(\mathrm{cos}x)}{dx}}=-\mathrm{sin}x$

Now,

$\frac{d(\mathrm{cos}x)}{dx}}\text{}={\displaystyle \frac{dt}{dx}$

$-\mathrm{sin}x={\displaystyle \frac{dt}{dx}}$

$-\mathrm{sin}xdx=dt$

Substitute $\mathrm{sin}xdx=-dt$ in the expression, we get

$\text{}\int (\sqrt{\mathrm{cos}x})sinxdx=-\int (\sqrt{t}\text{})dt$

We know, $\int {x}^{n}dx=\text{}{\displaystyle \frac{{x}^{n+1}}{n+1}}+C$

So, after integrating we get

$\text{}-\int (\sqrt{t}\text{})dt=-{\displaystyle \frac{{t}^{\frac{3}{2}}}{{\displaystyle \frac{3}{2}}}}+C$

$\text{}-\int (\sqrt{t}\text{})dt=-{\displaystyle \frac{2}{3}}\text{}{t}^{\frac{3}{2}}+C$

Re-substitute the value of t in terms of x, we get

$\text{}\int (\sqrt{\mathrm{cos}x})sinxdx=-{\displaystyle \frac{2}{3}}\text{}{\mathrm{cos}}^{\frac{3}{2}}x+C$

Note: Whenever an integrating expression consists of more than one function convert it into standard form by reduction method of integration such as substitution method of integration

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