Question

Integrate sinx+cosx/9+16sin2x from limit 0 to pi/4

Answer 1

this integration can be solved easily with help of substitution concept.

first resolve denominator of given term,

we know, sin2Φ = 2sinΦ.cosΦ

and sin²Φ+ cos²Φ = 1

so, sin2x = 2sinx.cosx

= 1 - 1 + 2sinx.cosx

= 1 -{1 - 2sinx.cosx }

= 1 - {sin²x + cos²x - 2sinx.cosx}

= 1 - (sinx - cosx)² ........(1)

now let sinx - cosx = r

differentiating both sides

or, (sinx + cosx)dx = dr.......(2)

now putting equations (1) and (2),

we get,

$\int{\frac{dr}{9+16(1-r^2)}}$

= $\int{\frac{1}{(5)^2-(4r)^2}}\,dr$

= $\frac{1}{2\times5\times4}log\left|\frac{5+4r}{5-4r}\right|$

now putting limits

r = (sinx - cosx)

lower limits , r = (sin0 + cos0)= -1

upper limits , r = (sinπ/4 - cosπ/4) = 0

so, $\left[\frac{1}{40}log\left|\frac{5+4r}{5-4r}\right|\right]^0_{-1}$

= 1/40 log[(5)/(5 )] - 1/40 log[(5 - 4)/(5 + 4)]

= 0 - 1/40 log[1/9]

= 1/40 log9

hence, answer is 1/40log9