Mathematics

$$\displaystyle \int\frac{\sin x.\cos x}{1+\sin^{4}x}d_{X}=$$


ANSWER

$$\displaystyle \frac{1}{2}\tan^{-1}(\sin^{2}x)+c$$


SOLUTION
=$$\displaystyle \frac {1}{2}\int \frac {\sin 2x}{1+\sin^4x}dx$$
=$$\displaystyle \frac {1}{2}\int \frac {d \sin^2x}{1+(\sin^2x)^2}=\frac {1}{2} \tan^{-1}(\sin^2x)+c$$
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 105
Enroll Now For FREE

Realted Questions

Q1 Subjective Medium
$$\int_{ - 1}^2 {\sqrt {5x + 6} } dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Single Correct Medium
$$\displaystyle \int \dfrac{1}{3\sin x +4\cos x} dx$$ equal, where $$\tan \alpha = \dfrac{4}{3}$$
  • A. $$\dfrac{1}{5} ln \left|\tan \left(x+\alpha\right)\right|+K$$
  • B. $$\dfrac{1}{5} ln \left|\tan \left(\dfrac{x}{2} +\dfrac{\pi}{2}- \dfrac{\alpha}{2}\right)\right|+K$$
  • C. $$\dfrac{1}{5} ln \left|\tan \left(\dfrac{x}{2} -\dfrac{\alpha}{2}\right)\right|+K$$
  • D. $$\dfrac{1}{5} ln \left|\tan \left(\dfrac{x}{2} + \dfrac{\alpha}{2}\right)\right|+K$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Single Correct Hard
Solve $$\displaystyle \int _{ 0 }^{ 1/2 }{ \dfrac { x\sin ^{ -1 }{ x }  }{ \sqrt { 1-{ x }^{ 2 } }  } dx= }$$
  • A. $$\dfrac { 1 }{ 2 } -\dfrac { \sqrt { 3 } \pi }{ 12 }$$
  • B. $$\dfrac { 1 }{ 2 } -\dfrac { \sqrt { 2 } \pi }{ 12 }$$
  • C. $$None\ of\ these$$
  • D. $$\dfrac { 1 }{ 2 } +\dfrac { \sqrt { 3 } \pi }{ 12 }$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Medium
Evaluate the following integral:

$$\displaystyle \int { \cfrac { { e }^{ x } }{ { e }^{ 2x }+5{ e }^{ x }+6 }  } dx\quad $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Passage Hard
Let us consider the integral of the following forms
$$f{(x_1,\sqrt{mx^2+nx+p})}^{\tfrac{1}{2}}$$
Case I If $$m>0$$, then put $$\sqrt{mx^2+nx+C}=u\pm x\sqrt{m}$$
Case II If $$p>0$$, then put $$\sqrt{mx^2+nx+C}=u\pm \sqrt{p}$$
Case III If quadratic equation $$mx^2+nx+p=0$$ has real roots $$\alpha$$ and $$\beta$$, then put $$\sqrt{mx^2+nx+p}=(x-\alpha)u\:or\:(x-\beta)u$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer