मराठी

∫ Cos 5 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int \cos^5 x\ dx\]
बेरीज
Advertisements

उत्तर

\[\text{ Let  I }= \int \cos^5 x \text{ dx }\]
\[ = \int \cos^4 x \cdot \text{ cos x dx}\] 
\[ = \int \left( \cos^2 x \right)^2 \text{ cos x dx} \]
\[ = \int \left( 1 - \sin^2 x \right)^2 \text{ cos x dx}\]
\[\text{ Putting  sin x = t}\]
\[ \Rightarrow \text{ cos x dx} = dt\]
\[ \therefore I = \int \left( 1 - t^2 \right)^2 \cdot dt\]
\[ = \int\left( t^4 - 2 t^2 + 1 \right) dt\]
\[ = \int t^4 \cdot dt - 2\int t^2 dt + \int dt\]
\[ = \frac{t^5}{5} - 2 \times \frac{t^{2 + 1}}{2 + 1} + t + C\]
\[ = \frac{t^5}{5} - \frac{2}{3} t^3 + t + C\]
\[ = \frac{\sin^5 x}{5} - \frac{2}{3} \sin^3 x + \sin x + C ........\left[ \because t = \sin x \right]\]

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 39 | पृष्ठ २०३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\int\left( 3x\sqrt{x} + 4\sqrt{x} + 5 \right)dx\]

\[\int\left( 2^x + \frac{5}{x} - \frac{1}{x^{1/3}} \right)dx\]

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

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

\[\int \left( 3x + 4 \right)^2 dx\]

\[\int \cos^{- 1} \left( \sin x \right) dx\]

If f' (x) = x + bf(1) = 5, f(2) = 13, find f(x)


` ∫   cos  3x   cos  4x` dx  

\[\int\frac{\sin 2x}{\sin 5x \sin 3x} dx\]

` ∫  tan 2x tan 3x  tan 5x    dx  `

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

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

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

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

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

` ∫  tan^3    x   sec^2  x   dx  `

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

\[\int \cos^7 x \text{ dx  } \]

\[\int\frac{1}{\sin^4 x \cos^2 x} dx\]

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

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

\[\int\frac{x^2}{x^2 + 7x + 10} dx\]

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

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

\[\int \log_{10} x\ dx\]

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

\[\int x^2 \sqrt{a^6 - x^6} \text{ dx}\]

\[\int\sqrt{3 - x^2} \text{ dx}\]

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

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

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

\[\int\frac{\sin^6 x}{\cos^8 x} dx =\]

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

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


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

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

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


\[\int\frac{\sin^6 x}{\cos x} \text{ dx }\]

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×