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If  \int \frac{3sinx+2cosx}{3cosx+2sinx}dx=ax+b\, ln\left |2sinx+3cosx \right |+C then

  • Option 1)

    a=-\frac{12}{13}\: ,b=\frac{15}{39}

  • Option 2)

    a=-\frac{17}{13}\: ,b=\frac{6}{13}

  • Option 3)

    a=-\frac{12}{13}\: ,b=-\frac{15}{39}

  • Option 4)

    none of these

 

Answers (1)

best_answer

As we learnt

Type of Integration by perfect square -

Integrals in the form of     \int \frac{p\cos x+q\sin x }{a\cos x+b\sin x}dx

- wherein

Working rule :

p\cos x+q\sin x=A\cdot \frac{\mathrm{d} }{\mathrm{d} x}(a\cos x+b\sin x)+B(a\cos x+b\sin x) 

Find A and B by comparing sinx and cosx 

 

 \int {\frac{{3\sin x + 2\cos x}}{{3\cos x + 2\sin x}}\;dx} =ax + b\ln |2\sin x + 3\cos x| + \;C

Diff. both sides, we get

\frac{{3\sin x + 2\cos x}}{{3\cos x + 2\sin x}} = a + \frac{{(2\cos x + 3\sin x)}}{{(2\sin x + 3\cos x)}}

=\frac{{\sin x \cdot (2a - 3b) + \cos x \cdot (3a + 2b)}}{{(3\cos x + 2\sin x)}}$

Comparing like terms on both sides, we get 3 = 2a – 3b, 2 = 3a + 2b

Þ        a = \frac{{12}}{{13}}\;,\,b = - \frac{{15}}{{39}}$.


Option 1)

a=-\frac{12}{13}\: ,b=\frac{15}{39}

Option 2)

a=-\frac{17}{13}\: ,b=\frac{6}{13}

Option 3)

a=-\frac{12}{13}\: ,b=-\frac{15}{39}

Option 4)

none of these

Posted by

gaurav

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