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maybe lol ..So that we can cancel out sec² Ө even before we begin integrating!
1 + tan² Ө = sec² Ө ; ∫1/(1 + x²) dx ; dx = sec² Ө . dӨ............... [do you see it now?
∫1/(1 + x²) dx
suppose x = tan Ө so Ө = tan^(-1) x
so 1 + x² = 1 + tan² Ө = sec² Ө
and dx/dӨ = sec² Ө
dx = sec² Ө . dӨ
So ∫1/(1 + x²) dx = ∫(1/sec² Ө) . sec² Ө dӨ
= ∫ 1 dӨ
= Ө + k
= tan^(-1) x + k
In these argument questions treat them the same way you treat trigonometry angles!
Validity ?
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_31.pdf
10 b!! please help me out with this! it's urgent!
Which two words ?waoww. that's really helpful...
P.S. You know those two words that need to be here
Thank you!
Aplogies for the photo quality and awesome handwritinghttp://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_31.pdf
10 b!! please help me out with this! it's urgent!
Whats this ?Aplogies for the photo quality and awesome handwriting
You integrate by the formula ;
Area= u. v - integral of udash.v.dx
there's this method of recognising u and vdash.
" L A T E "
L for logarithm, A for airthematic, T for trignometric and E for exponential...
the letter that comes first is u and the other is vdash.
in this question, airthematic comes first, in " L A T E " and exponential second. So x square is u and e raised to the power -x is vdash.
you convert vdash into v by integrating it individually. And you differentiate u and get the value of udash. Once you've done that, record the values in the side of the page or somewhere else. In the solution, they're recorded in a box. Put these values into the formula of the Area: u.v - intetgral of udash.v.dx.
There you get another integration by parts as you get the value of 2xe^ -2 in the integral sign . That indicates you have to integrate it by parts again and repeat the method again on the thing given under the integral sign.
Then you apply the limits to the equation and get the answer.
The ones you told me not to say , especially after being helped
Hey thank you so much! this was extremely helpful! best of luck for your exam, it's going to be good inshAllahAplogies for the photo quality and awesome handwriting
You integrate by the formula ;
Area= u. v - integral of udash.v.dx
there's this method of recognising u and vdash.
" L A T E "
L for logarithm, A for airthematic, T for trignometric and E for exponential...
the letter that comes first is u and the other is vdash.
in this question, airthematic comes first, in " L A T E " and exponential second. So x square is u and e raised to the power -x is vdash.
you convert vdash into v by integrating it individually. And you differentiate u and get the value of udash. Once you've done that, record the values in the side of the page or somewhere else. In the solution, they're recorded in a box. Put these values into the formula of the Area: u.v - intetgral of udash.v.dx.
There you get another integration by parts as you get the value of 2xe^ -2 in the integral sign . That indicates you have to integrate it by parts again and repeat the method again on the thing given under the integral sign.
Then you apply the limits to the equation and get the answer.
something like modulus of ax shud be less than 1 in expansion (1+ax)^n
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