IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

versatile_boy

Let sinx-cosx=t..........(1)

Diff. wrt x

Then (cosx+sinx)dx=dt

Also squaring (1)

1-sin2x= ${t}^{2}$

=> sin2x=1- ${t}^{2}$ .

Substituting these values in

$\int [sin(x) + cos(x)] / [9 + 16sin2(x)]$

to get $\int\frac{dt}{9+16(1- {t}^{2})}$ $= \int\frac{dt}{25-16 {t}^{2}}=\frac{1}{16}\int\frac{dt}{({\frac{5}{4}})^{2}-{t}^{2}}$ $\frac{1}{16*2*\frac{5}{4}}log(\frac{\frac{5}{4}+t}{\frac{5}{4}-t})$ $=\frac{1}{40}log\frac{5+4t}{5-4t}$ $=\frac{1}{40}log(\frac{5+4sinx-4cosx}{5-4sinx+4cosx})$

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