Mathematics

Solve:
$$\displaystyle \int {\dfrac{{2 - 3\sin x}}{{{{\cos }^2}x}}} dx.$$


SOLUTION

$$ \int \dfrac{\left( 2-3sinx \right)}{co{{s}^{2}}x}dx $$

$$ =\int{\left( \dfrac{2}{{{\cos }^{2}}x}-\dfrac{3\sin x}{{{\cos }^{2}}x} \right)dx} $$

$$ =\int{2{{\sec }^{2}}xdx-\int{3\dfrac{\sin x}{\cos x}\dfrac{1}{\cos x}}dx} $$

$$ =2\int{{{\sec }^{2}}xdx-3\int{\sec x\tan xdx}} $$

$$ =2\operatorname{tanx}-3secx+C $$

Hence, this is the answer.
View Full Answer

Its FREE, you're just one step away


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

Realted Questions

Q1 One Word Medium
$$\displaystyle \int_{-1/2}^{1/2}\cos x\log \left ( \frac{1+x}{1-x} \right )dx= ?$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q2 Subjective Medium
Evaluate: $$\int \dfrac{(x(\pi +49))^{15/7}}{\pi^2(x^{\pi}+7)}dx$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q3 Single Correct Medium
If $$\displaystyle \int\dfrac{\cos^4 x}{\sin^2 x}dx = A \cot x + B \sin 2x + C\dfrac{x}{2} + D$$, then
  • A. $$A = -2, B = 1/4$$
  • B. $$B = 1/4, C = -3$$
  • C. None of these
  • D. $$B = -1/4, C = -3$$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q4 Subjective Hard
$$\int { \dfrac { 2 }{ { \left( { t }^{ 2 }-1 \right)  }^{ 2 }-{ \left( 2 \right)  }^{ 2 } }  } $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer
Q5 Subjective Medium
Evaluate $$\int { \sin ^{ 3 }{ x } \cos ^{ 2 }{ x } dx } $$

Asked in: Mathematics - Integrals


1 Verified Answer | Published on 17th 09, 2020

View Answer