हिंदी

∫ Cot 4 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int \cot^4 x\ dx\]
योग

उत्तर

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

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Revision Excercise [पृष्ठ २०३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Revision Excercise | Q 31 | पृष्ठ २०३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

\[\int\frac{x^6 + 1}{x^2 + 1} dx\]

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

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

\[\int\frac{\tan x}{\sqrt{\cos x}} dx\]

\[\int x^3 \sin x^4 dx\]

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

\[\int \sec^4 2x \text{ dx }\]

\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]

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

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

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

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


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

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

` ∫    sin x log  (\text{ cos x ) } dx  `

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

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

\[\int\left( x + 1 \right) \text{ log  x  dx }\]

\[\int \cos^3 \sqrt{x}\ dx\]

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

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

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

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

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

\[\int\left( x - 2 \right) \sqrt{2 x^2 - 6x + 5} \text{  dx }\]

\[\int\frac{x^2 + x - 1}{x^2 + x - 6} dx\]

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

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

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

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


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

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

\[\int\frac{x + 1}{x^2 + 4x + 5} \text{  dx}\]

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

\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]


\[\int \tan^3 x\ \sec^4 x\ dx\]

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

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

Find: `int (sin2x)/sqrt(9 - cos^4x) dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×