C4 Integration by substitution question
Question is :
Use the substitution x=sinθ to show that, for x≤1,
The integral of 1/((1-x^2)^3/2) dx = x/((1-x^2)^1/2) +c
Where c is an arbitrary constant.
I got as far as substituting the x=sinθ into equation .
Giving
1/((1-sinθ^2)^3/2) dx
But where do I go from there? (since they still want it in terms of x at the end)
Thanks
Use the substitution x=sinθ to show that, for x≤1,
The integral of 1/((1-x^2)^3/2) dx = x/((1-x^2)^1/2) +c
Where c is an arbitrary constant.
I got as far as substituting the x=sinθ into equation .
Giving
1/((1-sinθ^2)^3/2) dx
But where do I go from there? (since they still want it in terms of x at the end)
Thanks
Original post by Muttley79
You need to substitute for dx as well.
Original post by the bear
and you need to recognize the integral of
1/cos2θ
1/cos2θ
When I came to marking my work afterwards,
the mark scheme said this :
∫ cosθ/cos^3θ dθ = ∫sec^2θ dθ and so on.
My question is: how did they get to ∫ cosθ/cos^3θ dθ from ∫ 1/((1-sin^2θ)^3/2) dθ .
I know that cos^2θ=1-sin^2θ so I could sub that in but how did they then get the cosθ on the top?
Thanks
Original post by EmEternal2
When I came to marking my work afterwards,
the mark scheme said this :
∫ cosθ/cos^3θ dθ = ∫sec^2θ dθ and so on.
My question is: how did they get to ∫ cosθ/cos^3θ dθ from ∫ 1/((1-sin^2θ)^3/2) dθ .
I know that cos^2θ=1-sin^2θ so I could sub that in but how did they then get the cosθ on the top?
Thanks
the mark scheme said this :
∫ cosθ/cos^3θ dθ = ∫sec^2θ dθ and so on.
My question is: how did they get to ∫ cosθ/cos^3θ dθ from ∫ 1/((1-sin^2θ)^3/2) dθ .
I know that cos^2θ=1-sin^2θ so I could sub that in but how did they then get the cosθ on the top?
Thanks
What did you get for dx?
Original post by Muttley79
What did you get for dx?
I got x=sinθ
so dx/dθ=cosθ
and dx=cosθ dθ
Original post by Muttley79
So substitute this for dx ...
Oh I see now, thanks, I'd got confused by the cosθ/cos^3θ but I realised that simplified down to 1/cos^2θ which is the sec^2θ they mentioned.
Thanks for your help!
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