Please kindly help, paper only award 4 marks so there has to be an easier solution. Wolfram use integration by parts which we are not taught. Thanks jt :)
Integrate from pi/16 to 0, [ 2 / (1 + sin 4x) ] + [ 2 / ( 1 - sin 4x) ] dx
Answer is 1.
heys, here's my attempt:
[ 2 / (1 + sin 4x) ] + [ 2 / ( 1 - sin 4x) ] dx
= 2 [(1-sin 4x + 1 + sin 4x)/(1-sin^2 4x)] dx
(factorise out the 2 on the numerator first, then combine the two fractions together, denominator simplified using (a+b)(a-b)=(a^2 - b^2))
= 4 [1/(1-sin^2 4x)] dx
yea. i'm not sure what A math scope is so, if there's a formula in your book that somehow corresponds to the expression then please use it. sorry that's all i can help.
Originally posted by Charmy:heys, here's my attempt:
[ 2 / (1 + sin 4x) ] + [ 2 / ( 1 - sin 4x) ] dx
= 2 [(1-sin 4x + 1 + sin 4x)/(1-sin^2 4x)] dx
(factorise out the 2 on the numerator first, then combine the two fractions together, denominator simplified using (a+b)(a-b)=(a^2 - b^2))= 4 [1/(1-sin^2 4x)] dx
yea. i'm not sure what A math scope is so, if there's a formula in your book that somehow corresponds to the expression then please use it. sorry that's all i can help.
There is no need for you to apologise, as long as you have tried your very best to help out.
[ 2 / (1 + sin 4x) ] + [ 2 / ( 1 - sin 4x) ]
= [ 2(1 - sin 4x) / (1 + sin 4x)( 1 - sin 4x) ] + [ 2(1 + sin 4x) / (1 + sin 4x)(1 - sin 4x) ]
= [ 2 - 2sin 4x + 2 + 2sin 4x ] / (1 + sin 4x)( 1 - sin 4x)
= 4 / [ 1 - (sin 4x)^2 ]
= 4 / [ (sin4x)^2 + (cos4x)^2 - (sin4x)^2 ]
= 4 / (cos4x)^2
= 4 (sec4x)^2
Integrate this, and you will get tan 4x, and you should be able to continue from here.
Thanks once again Forbiddensinner. I also got as far as 4 [1/(1-sin^2 4x)] dx and hit a brick wall. Appreciate it. :)