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\int \frac{1-7cos^{2}x}{\sin^{7}x\cos ^{2}x}dx=\frac{f(x)}{\left ( \sin x \right )^{7}}+C\: \: ,then\: \: f(x)=

  • Option 1)

    sin x

  • Option 2)

    cos x

  • Option 3)

    tan x

  • Option 4)

    cot x 

 

Answers (1)

best_answer

As we learnt

Different cases of type of indefinite integration -

(sin^{m}x)\left ( cos^{n} x\right ) \therefore \int \left ( sin^{m}xcos^{n}x \right )dx

Case     m      n

I          Odd    Even

II         Even   Odd

III        Odd     Odd

IV        Even   Even

- wherein

Substitution

I    Put t=cosx

II    Put t=sinx

III   Put t=sinx or t=cosx

iV  If (m+n)<0  put  t=tan x  or, If (m+n)>0  use DeMoivre's Theorem.

 

 \int \frac{1-7cos^{2}x}{\sin^{7}x\cos ^{2}x}dx=\int \left ( \frac{\sec ^{2}x}{\sin^{7}x}-\frac{7}{\sin ^{7}x} \right )dx=\int \frac{\sec ^{2}x}{\sin^{7}x}dx-\int \frac{7}{\sin ^{7}x}dx= I_{1}-I_{2}

Now \: I_{1}=\int \frac{\sec ^{2}x}{\sin^{7}x}dx=\frac{\tan x}{\sin^{7}x}+7\int \frac{\tan x.\cos x}{sin^{8}x}dx=\frac{\tan x}{\sin^{7}x}+I_{2}\therefore I_{1}-I_{2}= \frac{\tan x}{\sin^{7}x}+C\: \: \therefore f(x)=\tan x


Option 1)

sin x

Option 2)

cos x

Option 3)

tan x

Option 4)

cot x 

Posted by

gaurav

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