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Integrate: 3sinx+2cosx/3cosx+2sinx

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Solution:

          \Rightarrow I=int (3sinx+2cosx/3cosx+2sinx)dx

                \ N^r=lambda (D^r)+mu (d.chspace0.2cmofhspace0.2cmD^r)

        herefore      3sinx+2cosx=lambda (3cosx+2sinx)+mu (-3sinx+2cosx)

             Comparing the coefficient of sinx and cosx on both sides , We get

        3=2lambda -3mu hspace0.5cmandhspace0.2cm2=3lambda +2mu \ \Rightarrow lambda =12/13hspace0.5cmandhspace0.2cmmu =-5/13

    Hence,         \Rightarrow I=12/13int dx -(5/13)int -3sinx+2cosx/3cosx+2sinxhspace0.2cmdx

                  \Rightarrow I=(12/13)x-(5/13)log left | 3cosx+2sinx ight |+c

 

 

Posted by

Deependra Verma

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