1nisahnt bhaiya...will wht i said work for the ist integration
for first one multiply and divide by sin(x+a-(x+b))
More Integration....
1) \int{\frac{{dx}}{{\sqrt{\cos^{3}(x+\alpha )\sin (x+\beta )}}}}
2) \int{\left({x^{7m}+x^{2m}+x^{m}}\right)}\left({2x^{6m}+7x^{m}+14}\right)^{1/m}dx
3) \int{\frac{{dx}}{{\sqrt x+\sqrt{1-x}}}}
(Same comments as the previous QOD)
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9 Answers
1for first one multiply and divide by sin(x+a-(x+b))
will dat do nishant bhiya
??
19for (3)
put x= asin2θ
thus the integral reduces to
1/\sqrt{a}\int dx /\sin \theta +\cos \theta
now substitute...sinθ = 2(tanθ/2)/1+tan2θ/2 and cosθ =(1-tan2θ/2)/(1+tan2θ/2)......then replace tan θ/2 = t
.......this reduces the integral to 1/\sqrt{a}\int dt/at^{2}+bt +c
...its hopefully done!
11I = ∫ (x 7m + x 2m +x m) (2x 6m + 7 x m+ 14) 1/m dx
I = ∫ (x 7m + x 2m +x m) (2x 7m + 7 x 2m+ 14 x m) 1/m /x dx
I = ∫ (x 7m-1 + x 2m-1 + x m-1) (2x 7m +7x 2m + 14x m ) 1/m dx ............(1)
Put 2x 7m +7x 2m + 14x m = t, then
14m (x 7m-1 + x 2m-1 + x m-1)dx = dt
Therefore, eqn (1) becomes,
I = ∫ t 1/m dt / 14m = [(1/14m) t (1/m)+1] / [(1/m) +1 ] +C
Therefore, I = [1 / 14(m+1)] (2x 7m +7x 2m + 14x m ) (m+1) / m
1
62@Rohit.. something like that works.. but make sure you have completed the proof...
1yippie i got answer for Q1..... or i hope so i got :P
given is
\int \frac{dx}{\sqrt{cos^{3}(x+\alpha )sin(x+\beta )}}
put
( x + α ) = t → dx = dt and (x + β) = t - (α - β)
now this gives
=\int \frac{dt}{\sqrt{cos^{3}t.sin(t-(\alpha -\beta )}}
=\int \frac{dt}{\sqrt{cos^{3}t.\left\{sint.cos(\alpha -\beta )-cost.sin(\alpha -\beta ) \right\}}}
=\int \frac{dt}{\sqrt{cos^{4}t.\left\{tant.cos(\alpha -\beta )-sin(\alpha -\beta ) \right\}}}
=\int \frac{sec^{2}tdt}{\sqrt{\left\{tant.cos(\alpha -\beta )-sin(\alpha -\beta ) \right\}}}
put tant.cos(\alpha -\beta )-sin(\alpha -\beta )=u \Rightarrow sec^{2}t.cos(\alpha -\beta ) = du
now integral reduces to
\frac{1}{cos(\alpha -\beta )}\int \frac{sec^{2}t.cos(\alpha -\beta)dt}{\sqrt{tant.cos(\alpha -\beta) - sin(\alpha -\beta)}}
\frac{1}{cos(\alpha -\beta )}\int \frac{du}{\sqrt{u}}
2{sec(\alpha -\beta )}\sqrt{u}=2sec(\alpha -\beta )\sqrt{tan(x+\alpha )cos(\alpha -\beta )-sin(\alpha -\beta )}+c
hence answer is
2sec(\alpha -\beta )\sqrt{tan(x+\alpha )cos(\alpha -\beta )-sin(\alpha -\beta )}+c