मराठी

∫ E X Sec X ( 1 + Tan X ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\text{ Let I } = \int e^x \sec x\left( 1 + \tan x \right)dx\]

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

\[\text{ Here, }f(x) = \text{ sec x Put e}^x f(x) = t\]

\[ \Rightarrow f'(x) = \sec x \tan x\]

\[\text{ let e}^x \sec x = t\]

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

\[ e^x \sec x + e^x \sec x \tan x = \frac{dt}{dx}\]

\[ \Rightarrow e^x \left( \sec x + \tan x \right)dx = dt\]

\[ \therefore \int e^x \left( \sec x + \sec x \tan  x  \right)dx = \int dt\]

\[ = t + C\]

\[ = e^x \sec x + C\]

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 6 | पृष्ठ १४३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

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

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

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

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

 


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

\[\int\frac{\cos x}{\cos \left( x - a \right)} dx\] 

\[\int\frac{e^{3x}}{e^{3x} + 1} dx\]

` ∫  {sec  x   "cosec " x}/{log  ( tan x) }`  dx


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

` ∫  tan^3    x   sec^2  x   dx  `

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

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

\[\int \sin^7 x  \text{ dx }\]

\[\int\frac{2x + 3}{\sqrt{x^2 + 4x + 5}} \text{ dx }\]

\[\int\frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{  dx }\]

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

\[\int \left( \log x \right)^2 \cdot x\ dx\]

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

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

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

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

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

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

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

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

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

If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then


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


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

\[\int x \sin^5 x^2 \cos x^2 dx\]

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

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

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

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

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


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

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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×