∫ Tan 3 2 X Sec 2 X D X - Mathematics

Question

\[\int \tan^3 \text{2x sec 2x dx}\]

Sum

Solution

` ∫ tan^3 \text{x 2x . sec (2x) dx}`
\[ = \int \tan^2 2x . \text{sec 2x tan 2x dx}\]
` ∫ ( \sec^2 \left( 2x \right) - 1 \right) \text{sec (2x) tan ( 2x ) dx `
\[\text{Let sec }\left( 2x \right) = t\]
` ⇒ sec ( 2x ) tan (2x) × 2 = {dt}/{dx} `
` ⇒ sec ( 2x ) tan (2x) dx = {dt}/{2} `
\[Now, \int \tan^3\text{ x 2x} . \text{sec} \left( \text{2x }\right)dx\]
\[ = \frac{1}{2}\int\left( t^2 - 1 \right) dt\]
\[ = \frac{1}{2}\left[ \frac{t^3}{3} - t \right] + C\]
\[ = \frac{1}{2} \left[ \frac{\sec^3 \left( 2x \right)}{3} - \text{sec}\left( \text{2x }\right) \right] + C\]
\[ = \frac{1}{6} \text{sec}^3 \left( \text{2x} \right) - \frac{\text{sec} \left( \text{2x }\right)}{2} + C\]

shaalaa.com

Indefinite Integral Problems

Is there an error in this question or solution?

Q 62 Q 61 Q 63

Chapter 19: Indefinite Integrals - Exercise 19.09 [Page 59]

APPEARS IN

RD Sharma Mathematics [English] Class 12

Chapter 19 Indefinite Integrals
Exercise 19.09 | Q 62 | Page 59

Video TutorialsVIEW ALL [1]

view
Video Tutorials For All Subjects
Indefinite Integral Problems

video tutorial00:39:37

RELATED QUESTIONS

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

\[\int\frac{\sin^3 x - \cos^3 x}{\sin^2 x \cos^2 x} dx\]

\[\int\frac{5 \cos^3 x + 6 \sin^3 x}{2 \sin^2 x \cos^2 x} dx\]

\[\int \left( \tan x + \cot x \right)^2 dx\]

\[\int\frac{1 - \cos x}{1 + \cos x} dx\]

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

`∫ cos ^4 2x dx `

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

` ∫ {sin 2x} /{a cos^2 x + b sin^2 x } ` dx

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

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

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

\[\int\frac{1}{x^2 \left( x^4 + 1 \right)^{3/4}} dx\]

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

` ∫ \sqrt{tan x} sec^4 x dx `

\[\int \sin^7 x \text{ dx }\]

\[\int\frac{1}{\sin x \cos^3 x} dx\]

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

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

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

\[\int\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]

\[\int\frac{8 \cot x + 1}{3 \cot x + 2} \text{ dx }\]

`int"x"^"n"."log" "x" "dx"`

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

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

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

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

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

\[\int\frac{x^2 + 1}{x^4 + x^2 + 1} \text{ dx }\]

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

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

\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]

If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then

\[\int\frac{\sin 2x}{a^2 + b^2 \sin^2 x}\]

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

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

\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]

\[\int {cosec}^4 2x\ dx\]

\[\int\frac{1}{\sec x + cosec x}\text{ dx }\]

\[\int\frac{x + 3}{\left( x + 4 \right)^2} e^x dx =\]