Mathematics

The value of $$\displaystyle\int { \dfrac { \sqrt { \tan { x }  }  }{ \sin { x } \cos { x }  }  } dx$$ is equal to


ANSWER

$$2\sqrt { \tan { x } }+C $$


SOLUTION
Given $$\displaystyle\int \dfrac{\sqrt{\tan x}}{\sin x\cos x}$$
simplifying the function
$$=\displaystyle\int \dfrac{\sqrt{\tan x}}{\sin x.\cos x.\dfrac{\cos x}{\cos x}}$$
$$=\displaystyle\int\dfrac{\sqrt{\tan x}}{\sin x.\dfrac{\cos^{2}x}{\cos x}}$$
$$=\displaystyle\int\dfrac{\sqrt{\tan x}}{\cos^{2}x.\dfrac{\sin x}{\cos x}}$$
$$=\displaystyle\int\dfrac{\sqrt{\tan x}}{\cos^{2}x.\tan x}$$
$$=\displaystyle\int\dfrac{\sqrt{\tan x}.(\tan x)^{-1}}{\cos^{2}x}$$
$$=\displaystyle\int\dfrac{(\tan x)^{1/2^{-1}}}{\cos^{2}x}$$
$$=\displaystyle\int\dfrac{(\tan x)^{-1/2}}{\cos^{2}x}$$
$$=\displaystyle\int(\tan x)^{-1/2}.\dfrac{1}{\cos^{2}x}$$
$$=\displaystyle\int (\tan x)^{-1/2}.\sec^{2} x$$
Let $$\tan x=t$$
So, $$\sec^{2}x=\dfrac{dt}{dx}$$
$$\Rightarrow dx=\dfrac{dt}{\sec^{2}x}$$
$$\therefore \displaystyle\int (\tan x)^{-1/2}.\sec^{2}x.dx$$
$$=\displaystyle\int (t)^{-1/2}.\sec^{2}x.\dfrac{dt}{\sec^{2}x}$$
$$=\displaystyle\int (t)^{-1/2} dt$$
$$=\dfrac{t^{-1/2}+1}{-1/2 +1}+C$$      $$\left\{as \displaystyle\int x^{n}dx=\dfrac{x^{n+1}}{n+1}+C\right\}$$
$$=\dfrac{t^{1/2}}{1/2}+C$$
$$=2 t^{1/2}+C$$
$$=2\sqrt{t}+C$$
Substituting $$t=\tan x$$
$$=2\sqrt{\tan x}+C$$
View Full Answer

Its FREE, you're just one step away


Single Correct Medium Published on 17th 09, 2020
Next Question
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86
Enroll Now For FREE

Realted Questions

Q1 Single Correct Medium
$$\int \sqrt { 1 - \sin  x } d x =$$
  • A. $$2\sqrt {1-\sin x}+C$$
  • B. $$2\sqrt {1-2\sin x}+C$$
  • C. $$2\sqrt {1-\sin 2x}+C$$
  • D. $$2\sqrt {1+\sin x}+C$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Single Correct Medium

$$\displaystyle \int_{0}^{\pi/2}\frac{\cos ec^{2/3}x}{\cos ec^{2/3}x+\sec^{2/3}x}d_{X=}$$
  • A. $$\pi$$
  • B. $$-\pi$$
  • C. $$0$$
  • D. $$\pi/4$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Subjective Hard
Evaluate:
$$\int { \cfrac { { x }^{ 2 } }{ \left( 1+{ x }^{ 3 } \right)  }  } dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Medium
$$\int x^{3}d(tan^{-1}x)$$ is equal to

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Easy
Evaluate:
$$ \int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx} $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer