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\int \frac{sin^{3}x+cos^{3}x}{sin^{2}x\: cos^{2}x}dx=K_{1}\sec x+K_{2}\: cosecx +C, then which of the following is correct?

  • Option 1)

    K1 = K­2 = 1

  • Option 2)

    K1 = -K2 = 1

  • Option 3)

    K1 = K2 = -1

  • Option 4)

    K2 = 1 and K1 = -1

 

Answers (1)

As we learnt

Type of integration by substitution -

\int (f(x))^{n}\cdot f{}'(x)dx

\therefore \frac{\left [ f(x) \right ]^{n+1}}{n+1}+c

- wherein

Let f(x)=t

f{}'(x)dx=dt

 

 \int \frac{\sin x}{\cos^{2}x}dx+\int \frac{\sin^{2}x}{\cos x}dx 

Putting cosx = t  and sin x = t  respectively.

\therefore \frac{-\left ( \cos x \right )^{-2+1}}{-2+1}+ \frac{\left ( \sin x \right )^{-2+1}}{-2+1}+c

\therefore \sec x-cosec x+c        

Þ  K1 = 1,  K2 = -1

 


Option 1)

K1 = K­2 = 1

Option 2)

K1 = -K2 = 1

Option 3)

K1 = K2 = -1

Option 4)

K2 = 1 and K1 = -1

Posted by

Vakul

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