Question

solve the above attachment ​

Answer 1

Solution:

Given Integral:

$\displaystyle\rm\longrightarrow I=\int\dfrac{\sqrt{\tan(x)}}{\sin(x)\cos(x)}\: dx$

Can be written as:

$\displaystyle\rm\longrightarrow I=\int\dfrac{\sqrt{\tan(x)}}{\dfrac{\tan(x)}{\sec(x)}\cdot\dfrac{1}{\sec(x)}}\: dx$

$\displaystyle\rm\longrightarrow I=\int\dfrac{\sec^{2}(x)\sqrt{\tan(x)}}{\tan(x)}\: dx$

$\displaystyle\rm\longrightarrow I=\int\dfrac{\sec^{2}(x)}{\sqrt{\tan(x)}}\: dx$

Now, let us assume that:

$\rm\longrightarrow u=\tan(x)$

$\rm\longrightarrow du=\sec^{2}(x)\:dx$

$\rm\longrightarrow dx=\dfrac{1}{\sec^{2}(x)}\:du$

Therefore, the integral becomes:

$\displaystyle\rm\longrightarrow I=\int\dfrac{1}{\sqrt{u}}\:du$

$\displaystyle\rm\longrightarrow I=2\sqrt{u}+C$

Substitute back u = tan(x), we get:

$\displaystyle\rm\longrightarrow I=2\sqrt{\tan(x)}+C$

Therefore:

$\displaystyle\rm\longrightarrow \int\dfrac{\sqrt{\tan(x)}}{\sin(x)\cos(x)}\:dx=2\sqrt{\tan(x)}+C$

Which is our required answer.

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$

Answer 2

Answer:

$\displaystyle\rm \int\dfrac{\sqrt{\tan x}}{\sin x \cos x}dx = 2\sqrt{\tan(x)} + c$

Step-by-step explanation:

We have,

${\displaystyle \longrightarrow\small\rm I = \int\dfrac{\sqrt{\tan x}}{\sin x \cos x}dx }$

Can be rewritten as,

${\displaystyle \longrightarrow\small\rm I = \int\dfrac{\sqrt{\tan x}}{\sin x \cos x\cdot\frac{\cos x}{\cos x}}dx }$

${\displaystyle \longrightarrow\small\rm I = \int\dfrac{\sqrt{\tan x}}{\tan x\cos^2x}dx }$

${\displaystyle \longrightarrow\small\rm I = \int\dfrac{\sec^2x}{\sqrt{\tan x}}dx }$

Now we will solve it further using the method of substitution.

$\rm Let \tan x = t \implies \sec^2x dx = dt$

By substituting, the given integral changes to:

${\displaystyle \longrightarrow\small\rm I = \int\dfrac{dt}{\sqrt{t}} }$

${\displaystyle \longrightarrow\small\rm I = \int t^\frac{-1}{2}dx }$

Now we will use the below power rule of integration.

• $\boxed{\tt\int x^n dx = \dfrac{x^{n+1}}{n+1}}$

Using the formula, we get:

${\displaystyle \longrightarrow\small\rm I = \dfrac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+ 1} + c }$

${\displaystyle \longrightarrow\small\rm I = \dfrac{t^{\frac{1}{2}}}{\frac{1}{2}} + c }$

${\displaystyle \longrightarrow\small\rm I = 2\sqrt{t}+ c }$

Substituting back the value of dummie variable t.

${\displaystyle \longrightarrow\small\rm I =\sqrt{\tan x} + c }$

Hence the required answer is:

$\underline{\boxed{\red{\tt \int\dfrac{\sqrt{\tan x}}{\sin x \cos x}dx = 2\sqrt{\tan(x)} + c}}}$

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