06-09-09 Integration

2nd sum is a previous year question:P

for first one multiply and divide by sin(x+a-(x+b))

will dat do nishant bhiya
??

I = ∫ (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}

yippie 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