मराठी

∫ Sec 2 X Cos 2 2 X Dx - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\int\left( \sec^2 x \cdot \cos^2 2x \right)dx\]
\[ = \int \sec^2 x \times \left( 2 \cos^2 x - 1 \right)^2 dx\]
\[ = \int \sec^2 x \left[ 4 \cos^4 x - 4 \cos^2 x + 1 \right]dx\]
\[ \Rightarrow \int\left( 4 \cos^2 x - 4 + \sec^2 x \right)dx\]
\[ = 4\int \cos^2 x \text{ dx } + \int \sec^2 x \text{ dx }- 4\int dx\]
\[ \Rightarrow 4\int\left( \frac{1 + \cos 2x}{2} \right)dx + \int \sec^2 x - 4\int dx\]
\[ \Rightarrow 2 \left[ x + \frac{\sin 2x}{2} \right] + \tan x - 4x + C\]
\[ \Rightarrow \sin 2x + \tan x - 2x + C\]

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 9 | पृष्ठ २०३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

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

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


\[\int\frac{1}{\left( 7x - 5 \right)^3} + \frac{1}{\sqrt{5x - 4}} dx\]

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

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

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

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

` ∫    x   {tan^{- 1} x^2}/{1 + x^4} dx`

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

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

\[\int\frac{1}{x^2 - 10x + 34} dx\]

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

\[\int\frac{1}{\sqrt{16 - 6x - x^2}} dx\]

\[\int\frac{e^x}{\sqrt{16 - e^{2x}}} dx\]

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

\[\int\frac{\left( 3 \sin x - 2 \right) \cos x}{5 - \cos^2 x - 4 \sin x} dx\]

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

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

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

\[\int x \text{ sin 2x dx }\]

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

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

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

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

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

\[\int\frac{1}{\cos x \left( 5 - 4 \sin x \right)} dx\]

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

\[\int\frac{x^2 + 9}{x^4 + 81} \text{ dx }\]

 


If \[\int\frac{\cos 8x + 1}{\tan 2x - \cot 2x} dx\]


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

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

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


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

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


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

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

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

\[\int\frac{\cos^7 x}{\sin x} dx\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×