Mathematics

Integrate: $$\int \dfrac{\sec^2\sqrt{x}}{\sqrt{x}}dx$$


ANSWER

$$I = 2 \tan \sqrt x+c$$


SOLUTION
$$I=\int { \frac { { sec }^{ 2 }\sqrt { x }  }{ \sqrt { x }  }  } dx\\ Let\quad u=\sqrt { x } \Rightarrow du=\frac { 1 }{ 2\sqrt { x }  } dx\\ $$
Substituting in I, 

$$I=2\int { { sec }^{ 2 }udu } \\        =2tanu+c\quad \quad \quad \left[ \because \frac { d\left( tanu \right)  }{ du } { sec }^{ 2 }u \right] $$

$$\Rightarrow I=2tan\sqrt { x } +c$$

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 84
Enroll Now For FREE

Realted Questions

Q1 Subjective Medium
Prove the following:
$$\displaystyle \int [f(x)]^{n}f'(x) dx = \dfrac {[f(x)]^{n + 1}}{n + 1} + C$$, if $$n\neq -1$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Subjective Hard
Solve:
$$\displaystyle \int \cfrac{x^{2}}{1+x^{4}} d x$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Single Correct Hard
Evaluate $$\displaystyle {\int \sin^{-1}\, \sqrt {\frac {x}{a\, +\, x}} dx}$$
  • A. $$(a + x) \tan^{-1} \displaystyle \sqrt \frac {x}{a}\, -\, \sqrt {ax}\, +\, c$$
  • B. $$a \tan^{-1} \displaystyle \sqrt \frac {x}{a}\, -\, \sqrt {ax}\, +\, c$$
  • C. $$(a + x) \tan^{-1} \displaystyle \sqrt {x}\, -\, \sqrt {ax}\, +\, c$$
  • D. $$\displaystyle \left(\frac { ax+1 }{ 2a } \right)\sin^{ -1 }\, \sqrt { \frac { x }{ a\, +\, x } } -\frac { 1 }{ 2 } \sqrt { \frac { x }{ a } } +c$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Medium
Evaluate the given integral
$$\int { x.cosec ^{ 2 }{ x }  } dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Medium
$$\int \left( {4x + 2} \right)\sqrt {{x^2} + x + 1} dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer