मराठी

∫ Tan 4 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int \tan^4 x\ dx\]
बेरीज
Advertisements

उत्तर

\[\text{ Let I } = \int \text{ tan}^4 \text{ x dx }\]
\[ = \int \tan^2 x \cdot \tan^2 \text{ x dx}\]
\[ = \int\left( \sec^2 x - 1 \right) \tan^2 \text{ x  dx}\]
\[ = \int \sec^2 x \cdot \tan^2\text{  x dx }- \int \tan^2 \text{ x  dx}\]
\[ = \int \tan^2 x \cdot \sec^2 x - \int\left( \sec^2 x - 1 \right) dx\]
\[\text{ Putting tan x } = \text{ t  in   the  Ist  integral}\]
\[ \Rightarrow \sec^2 \text{ x  dx } = dt\]
\[ \therefore I = \int t^2 \cdot dt - \int\left( \sec^2 x - 1 \right) dx\]
\[ = \frac{t^3}{3} - \tan x + x + C\]
\[ = \frac{\tan^3 x}{3} - \text{ tan x + x + C }..............\left( \because t = \tan x \right)\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Revision Excercise [पृष्ठ २०३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Revision Excercise | Q 29 | पृष्ठ २०३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

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

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

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

\[\int\frac{\text{sin }\left( \text{2 + 3 log x }\right)}{x} dx\]

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

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

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

\[\int\frac{\cos 2x}{\sqrt{\sin^2 2x + 8}} dx\]

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

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

`  ∫ \sqrt{"cosec x"- 1}  dx `

\[\int\frac{1 - 3x}{3 x^2 + 4x + 2}\text{  dx}\]

\[\int\frac{x + 2}{2 x^2 + 6x + 5}\text{  dx }\]

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

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

\[\int\frac{1}{5 + 4 \cos x} dx\]

\[\int x^2 \cos 2x\ \text{ dx }\]

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

\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]

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

\[\int\frac{1}{x\left[ 6 \left( \log x \right)^2 + 7 \log x + 2 \right]} dx\]

\[\int\sqrt{\cot \text{θ} d  } \text{ θ}\]

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

Write the anti-derivative of  \[\left( 3\sqrt{x} + \frac{1}{\sqrt{x}} \right) .\]


\[\int x^{\sin x} \left( \frac{\sin x}{x} + \cos x . \log x \right) dx\] is equal to

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

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

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

\[\int\text{ cos x  cos  2x   cos  3x  dx}\]


\[\int \cot^4 x\ dx\]

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

\[\int\frac{1}{3 x^2 + 13x - 10} \text{ dx }\]

\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]

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

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

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×