IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

richeek

integrate((1/2sinx+cosx+3)dx)
sorry couldnt use the symbols!

visioniit

itz of the type integral of 1/a sinx+b cosx +c..
itz general method..

magiclko

its 1 / (2sin x) or (1/2) sin x???

nitin62225

convert sinx , cos x into tan (x/2)

sin x= 2tan(x/2)/sec²(x/2) , cos x= (1-tan²(x/2))/sec²(x/2)

then put tan(x/2)=t, solve further......

iitkgp_bipin

Integral is of the form : I =

dx/(2sinx + cosx + 3)

Substitute cosx = (1 - tan²x/2)/(1 + tan²x/2) and sinx = (2tanx/2)/(1 + tan²x/2) in the above integral.
The numerator shall become sec²x/2 and the denominator will be a quadratic in tanx/2.

Now put tanx/2 = t i.e. sec²x/2 = 2dt.

The integral will become I =

2.dt/{2(2t) + (1-t²) + 3}

I =

2.dt/(4 + 4t - t²)

Now rest of the part is easy and at last back-substitute tanx/2 = t.

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