मराठी

∫ Tan X √ Cos X D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\int\frac{\tan x}{\sqrt{\cos x}}dx\]
\[ \Rightarrow \int\frac{\sin x}{\cos x \sqrt{\cos x}} dx\]
\[ \Rightarrow \int\frac{\sin x}{\cos {}^\frac{3}{2} x}dx\]
\[Let \cos x = t\]
\[ \Rightarrow - \text{sin x dx }= dt\]
\[ \Rightarrow \sin x = - \frac{dt}{dx}\]
\[Now, \int\frac{\sin x}{\cos {}^\frac{3}{2} x}dx\]


\[ = \int - \frac{1}{t^\frac{3}{2}}dt\]

 


\[ = - \int t^{- \frac{3}{2}} dt\]

 


\[ = - \left[ \frac{t^{- \frac{3}{2} + 1}}{\frac{- 3}{2} + 1} \right] + C\]
\[ = \frac{2}{\sqrt{t}} + C\]
\[ = \frac{2}{\sqrt{\cos x}} + C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.09 [पृष्ठ ५८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.09 | Q 12 | पृष्ठ ५८

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

\[\int \tan^2 \left( 2x - 3 \right) dx\]


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

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

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

` ∫  sec^6   x  tan    x   dx `

\[\int {cosec}^4  \text{ 3x } \text{ dx } \]

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

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

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

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

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

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

\[\int\frac{x^2 + 1}{x^2 - 5x + 6} dx\]

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

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

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

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

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

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

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

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

\[\int\frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} dx\]

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

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

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

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

\[\int\frac{\sin x}{3 + 4 \cos^2 x} dx\]

\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then 


\[\int\frac{x^3}{x + 1}dx\] is equal to

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

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

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

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

\[\int \tan^5 x\ \sec^3 x\ dx\]

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

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

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

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×