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  • Integrate the functions in Exercises 1 to 24. 10.

Integrate the functions in Exercises 1 to 24.

    Q10.    \frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}

Answers (1)

best_answer

We have         

                                      I\ =\ \int \frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}

Simplifying the given expression, we get :

               

                                   \frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}\ =\ \frac{(\sin^4x + \cos^4x)(\sin^4x - \cos^4x) }{1- 2\sin^ x\cos^2 x}

or                                                                     =\ \frac{(\sin^4x + \cos^4x)(\sin^2x - \cos^2x)(\sin^2x + \cos^2x) }{1- 2\sin^ x\cos^2 x}

or                                                                     =\ -\frac{(\sin^4x + \cos^4x)(\cos^2x - \sin^2x) }{1- 2\sin^ x\cos^2 x}

or                                                                      =\ -\cos^2x - \sin^2x\ =\ -\cos 2x

Thus,                                

                     I\ =\ \int \frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}\ =\ -\int \cos 2x\ dx

and                                                                       =\ - \frac{\sin 2x}{2}\ +\ C

 

Posted by

Devendra Khairwa

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