English

∫ ( Sec 2 X + C O S E C 2 X ) D X - Mathematics

Advertisements
Advertisements

Question

\[\int\left( \sec^2  x + {cosec}^2  x \right)  dx\]
Sum
Advertisements

Solution

\[\int\left( \sec^2 x + {cosec}^2 x \right)dx\]
\[ = \int \sec^2\text{ x    dx} + \int {cosec}^2\text{ x  dx}\]
\[ = \tan x - \cot x + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.02 [Page 15]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.02 | Q 22 | Page 15

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

\[\int \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 dx\]

If f' (x) = 8x3 − 2xf(2) = 8, find f(x)


\[\int\frac{\sec^2 x}{\tan x + 2} dx\]

\[\int\left( 4x + 2 \right)\sqrt{x^2 + x + 1}  \text{dx}\]

\[\int\frac{\sin 2x}{\left( a + b \cos 2x \right)^2} dx\]

\[\int\frac{\sin\sqrt{x}}{\sqrt{x}} dx\]

\[\int \sin^5 x \text{ dx }\]

` = ∫1/{sin^3 x cos^ 2x} dx`


\[\int\frac{1}{\sqrt{2x - x^2}} dx\]

\[\int\frac{1}{\sqrt{7 - 3x - 2 x^2}} dx\]

\[\int\frac{1}{\sqrt{5 x^2 - 2x}} dx\]

\[\int\frac{1}{x\sqrt{4 - 9 \left( \log x \right)^2}} dx\]

\[\int\frac{\sin 2x}{\sqrt{\sin^4 x + 4 \sin^2 x - 2}} dx\]

\[\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 - \cos^2 x - 7\sin x}dx\]

\[\int\frac{x + 1}{\sqrt{x^2 + 1}} dx\]

\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]

\[\int\frac{3 + 2 \cos x + 4 \sin x}{2 \sin x + \cos x + 3} \text{ dx }\]

\[\int\frac{1}{4 + 3 \tan x} dx\]

\[\int \sec^{- 1} \sqrt{x}\ dx\]

\[\int\frac{\sin^{- 1} x}{x^2} \text{ dx }\]

\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]

\[\int e^x \left[ \sec x + \log \left( \sec x + \tan x \right) \right] dx\]

\[\int e^x \frac{1 + x}{\left( 2 + x \right)^2} \text{ dx }\]

\[\int\sqrt{x^2 - 2x} \text{ dx}\]

\[\int\frac{x^2 + 1}{x\left( x^2 - 1 \right)} dx\]

\[\int\frac{5 x^2 - 1}{x \left( x - 1 \right) \left( x + 1 \right)} dx\]

\[\int\frac{18}{\left( x + 2 \right) \left( x^2 + 4 \right)} dx\]

Evaluate the following integral:

\[\int\frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)}dx\]

\[\int\frac{\left( x - 1 \right)^2}{x^4 + x^2 + 1} \text{ dx}\]

\[\int\frac{1}{\left( x + 1 \right) \sqrt{x^2 + x + 1}} \text{ dx }\]

\[\int\frac{x}{4 + x^4} \text{ dx }\] is equal to

If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then


\[\int \sin^4 2x\ dx\]

\[\int \tan^{- 1} \sqrt{x}\ dx\]

\[\int\frac{e^{m \tan^{- 1} x}}{\left( 1 + x^2 \right)^{3/2}} \text{ dx}\]

\[\int\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} \text{ dx}\]

\[\int\frac{1}{1 + x + x^2 + x^3} \text{ dx }\]

\[\int\frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} \text{ dx}\]

Find : \[\int\frac{dx}{\sqrt{3 - 2x - x^2}}\] .


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×