indefinite integral

sharrmad prabhu verlekar

integral sin(x)cos(x)/a^2cos^2(x)+b^2sin^2(x)

ROHAN KUMAR

2SinxCosx=Sin2x now differentating a^2cos^2x + b^2sin^2x u will get ( b^2 -a^2)sin2x ................................... So take a^2cos^2x+b^2sin^2x=t dt=written above ................................ So doing this substitution u can easily solve this problem

Manasi

sinx cosx dx/ [a²cos²x + b²sin²x]

=> sin2x dx/2 [a²cos²x + b²sin²x] .............(A)

Let [a²cos²x + b²sin²x] = t

Differentiating both the sides, we have

sin 2x [b²-a²] dx = dt

=> sin 2x dx = dt / [b²-a²]

Substituting in equation (A)

integral of dt /{2t [b²-a²] }

=> ln (t) / {2 [b²-a²]}

substitute the value of 't' and get the answer :)

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