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Question

Integrate 1sinx-2cosx2sinx+cosxdx


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Solution

Step:1 Simplify the given integral

Given: 1sinx-2cosx2sinx+cosxdx

We can simplify it as,

=12sin2x+sinxcosx-4sinxcosx-2cos2xdx

Dividing the numerator and denominator by cos2x,

Now,

=1cos2x2sin2x-3sinxcosx-2cos2xcos2xdx

=sec2x2tan2x-3tanx-2dx

Step: 2 Using the method of substitution to find the integral

Put tanx=t so that sec2xdx=dt

Now,

sec2x2tan2x-3tanx-2dx

=dt2t2-3t-2=121t2-3t4-1dt=121t-342-522dt

Step: 3 Using the appropriate formula to find the value of integral

We know that,

1x2-a2dx=12alogx-ax+a+C

Now,

=12×1254logt-34-54t-34+54+C=15logt-2t+12+C=15log2tanx-42tanx+1+C

Hence, Integral 1sinx-2cosx2sinx+cosxdx is15log2tanx-42tanx+1+C, where C is an arbitrary constant.


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