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  • Evaluate the following: int root tanx dx (Hint: Put tan x = t2)

Evaluate the following:

\int \sqrt{\tan x} d x (Hint: Put tan x = t^2)

Answers (1)

Given:

\int \sqrt{\tan x} d x

Put tan x = t^2

\begin{aligned} &\Rightarrow \sec ^{2} x d x=2 t d t\\ &\Rightarrow \mathrm{dx}=\frac{2 t \mathrm{dt}}{\sec ^{2} x}=\frac{2 \mathrm{tdt}}{1+\tan ^{2} \mathrm{x}}=\frac{2 \mathrm{tdt}}{1+\mathrm{t}^{4}}\\ &\Rightarrow \int \sqrt{\tan \mathrm{x}} \mathrm{d} \mathrm{x}=\int \sqrt{\mathrm{t}^{2}} \frac{2 \mathrm{t} \mathrm{dt}}{1+\mathrm{t}^{4}}=\int \frac{2 \mathrm{t}^{2} \mathrm{dt}}{1+\mathrm{t}^{4}}\\ &\Rightarrow \int \frac{2 t^{2} d t}{1+t^{4}}=\int \frac{\left(2 t^{2}+1-1\right) d t}{1+t^{4}}=\int \frac{\left(t^{2}+1\right)+\left(t^{2}-1\right)}{1+t^{4}} d t\\ &\Rightarrow \int \frac{\left(t^{2}+1\right)+\left(t^{2}-1\right)}{1+t^{4}} d t=\int \frac{\left(t^{2}+1\right)}{1+t^{4}} d t+\int \frac{\left(t^{2}-1\right)}{1+t^{4}} d t \end{aligned}

Taking out t^2 common in both the numerators

\\ \Rightarrow \int \frac{\left(t^{2}+1\right)}{1+t^{4}} d t+\int \frac{\left(t^{2}-1\right)}{1+t^{4}} d t=\int \frac{t^{2}\left(1+\left(\frac{1}{t^{2}}\right)\right)}{1+t^{4}} d t+\int \frac{t^{2}\left(1-\left(\frac{1}{t^{2}}\right)\right)}{1+t^{4}} \mathrm{dt} \\ \int \frac{t^{2}\left(1+\left(\frac{1}{t^{2}}\right)\right)}{1+t^{4}} \mathrm{dt}+\int \frac{t^{2}\left(1-\left(\frac{1}{t^{2}}\right)\right)}{1+t^{4}} \mathrm{dt}=\int \frac{\left(1+\left(\frac{1}{t^{2}}\right)\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt}+\int \frac{\left(1-\left(\frac{1}{t^{2}}\right)\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt} \ldots . \text { (1) }

\\ \begin{aligned} &\Rightarrow \operatorname{Now} t^{2}+\frac{1}{t^{2}}=\left(t \pm \frac{1}{t}\right)^{2} \mp 2 \ldots(3)\\ &=\operatorname{for}(a) \int \frac{\left(1+\left(\frac{1}{t^{2}}\right)\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt} \text { taking } t-\frac{1}{t}=z\\ &\Rightarrow\left(1+\frac{1}{t^{2}}\right) d t=d z \end{aligned}

\\ =\int \frac{\left(1+\frac{1}{t^{2}}\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt}=\int \frac{\left(1+\frac{1}{t^{2}}\right)}{\left(t-\frac{1}{t}\right)^{2}+2} \mathrm{dt} \\ =\int \frac{\left(1+\frac{1}{t^{2}}\right)}{\left(t-\frac{1}{t}\right)^{2}+2} \mathrm{dt}=\int \frac{\mathrm{d} z}{z^{2}+2}=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{z}{\sqrt{2}}\right)+\mathrm{c} \ldots(2)

\\ \begin{aligned} &\text { for (b) } \int \frac{\left(1-\left(\frac{1}{t^{2}}\right)\right)}{\frac{1}{t^{2}}+t^{2}} \text { dt taking } t+\frac{1}{t}=z\\ &\Rightarrow\\ &\Rightarrow\left(1-\frac{1}{t^{2}}\right) \mathrm{dt}=\mathrm{dz}\\ &=\int \frac{\left(1-\frac{1}{t^{2}}\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt}=\int \frac{\left(1-\frac{1}{\mathrm{t}^{2}}\right)}{\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)^{2}-2} \mathrm{dt}\\ &=\int \frac{\left(1-\frac{1}{t^{2}}\right)}{\left(t+\frac{1}{t}\right)^{2}-2} \mathrm{dt}=\int \frac{\mathrm{d} z}{z^{2}-2}=\frac{1}{2 \sqrt{2}} \ln \left|\frac{z-\sqrt{2}}{z+\sqrt{2}}\right|+\mathrm{c} \ldots(3) \end{aligned}

 

 

Put (2) and (3) in (1)

\\ =\int \frac{\left(1+\left(\frac{1}{t^{2}}\right)\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt}+\int \frac{\left(1-\left(\frac{1}{t^{2}}\right)\right)}{\frac{1}{t^{2}}+t^{2}} \mathrm{dt}=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\left(t-\frac{1}{t}\right)}{\sqrt{2}}\right)+\frac{1}{2 \sqrt{2}} \ln \left|\frac{\left(t+\frac{1}{\mathrm{t}}\right)-\sqrt{2}}{\left(\mathrm{t}+\frac{1}{\mathrm{t}}\right)+\sqrt{2}}\right|+\mathrm{c} \\ \Rightarrow \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\mathrm{t}^{2}-1}{\mathrm{t} \sqrt{2}}\right)+\frac{1}{2 \sqrt{2}} \ln \left|\frac{\left(\mathrm{t}^{2}+1-\mathrm{t} \sqrt{2}\right.}{\left(\mathrm{t}^{2}+1+\mathrm{t} \sqrt{2}\right.}\right|+\mathrm{C}

\\ \Rightarrow \text{Now again putting} t=\sqrt{\tan x}\text{ to obtain the final result} \\ =\int \sqrt{\tan x} d x=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\tan x-1}{\sqrt{2 \tan x}}\right)+\frac{1}{2 \sqrt{2}} \ln \left|\frac{(\tan x+1-\sqrt{2 \tan x}}{\tan x+1+\sqrt{2 \tan x}}\right| \\

 

 

 

 

 

Posted by

infoexpert22

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