English

∫ X Sin − 1 X 2 √ 1 − X 4 D X - Mathematics

Advertisements
Advertisements

Question

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

Solution

\[\int \frac{x \sin^{- 1} x^2}{\sqrt{1 - x^4}}dx\]
\[\text{Let} \sin^{- 1} x^2 = t\]
\[ \Rightarrow \frac{1 \times 2x}{\sqrt{1 - x^4}} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{x        dx}{\sqrt{1 - x^4}} = \frac{dt}{2}\]
\[Now, \int \frac{x \sin^{- 1} x^2}{\sqrt{1 - x^4}}dx\]
\[ = \frac{1}{2}\ ∫    tdt\]
\[ = \frac{t^2}{4} + C\]
\[ = \frac{\left( \sin^{- 1} x^2 \right)^2}{4} + C\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Indefinite Integrals - Exercise 19.09 [Page 58]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 19 Indefinite Integrals
Exercise 19.09 | Q 35 | Page 58

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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

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

\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]

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

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

` ∫  sec^6   x  tan    x   dx `

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

\[\int \cos^7 x \text{ dx  } \]

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

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

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

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

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

\[\int\frac{1}{1 - \cot x} dx\]

`int"x"^"n"."log"  "x"  "dx"`

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

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

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

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

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

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

\[\int\left( x - 1 \right) e^{- x} dx\] is equal to

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


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

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

The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to


\[\int\frac{\sin 2x}{a^2 + b^2 \sin^2 x}\]

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

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

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

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

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

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


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

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

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

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

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×