मराठी

∫ { Tan ( Log X ) + Sec 2 ( Log X ) } D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\left\{ \tan \left( \log x \right) + \sec^2 \left( \log x \right) \right\} dx\]
बेरीज
Advertisements

उत्तर

\[\text{ Let I } = \int\left[ \tan\left( \log x \right) + \sec^2 \left( \log x \right) \right]dx\]

\[\text{ Put  log x = t }\]

\[ \Rightarrow x = e^t \]

\[ \Rightarrow dx = e^t dt\]

\[ \text{ ∴  I }= \int\left( \tan t + \sec^2 t \right) e^t dt\]

\[\text{ Here,} f(t) = \tan t\]

\[ \Rightarrow f'(t) = \sec^2 t\]

` \text{ let e}^t \tan(t) = p  `

\[\text{ Diff  both   sides  w . r . t t }\]

\[ e^t \left[ \tan t + \sec^2 t \right] = \frac{dp}{dt}\]

\[ \Rightarrow e^t \left[ \tan t + \sec^2 t \right]dt = dp\]

\[ ∴  I = \int dp\]

\[ = p + C\]

\[ = e^t \tan t + C\]

\[ = x \text{ tan (log x) }+ C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.26 [पृष्ठ १४३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.26 | Q 22 | पृष्ठ १४३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

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

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

Integrate the following integrals:

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

\[\int x^2 e^{x^3} \cos \left( e^{x^3} \right) dx\]

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

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

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

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

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

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

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

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

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

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

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

\[\int e^\sqrt{x} \text{ dx }\]

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

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

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

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

\[\int e^x \sec x \left( 1 + \tan x \right) dx\]

\[\int e^x \left( \cot x + \log \sin x \right) dx\]

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

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

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

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

If `int(2x^(1/2))/(x^2)  dx = k  .  2^(1/x) + C`, then k is equal to ______.


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

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

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


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

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


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


\[\int\sqrt{\frac{a + x}{x}}dx\]
 

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×