[sabre2th1]

For a) and b) are you expected to spot that the numerator is basically the denominator differentiated? Also, do you always use ln when thisis the case?

For c), how would you know what to let y = .. (they done sin³x, how would you know this?)

[raheem94]

(Original post by sabre2th1)


For a) and b) are you expected to spot that the numerator is basically the denominator differentiated? Also, do you always use ln when thisis the case?

For c), how would you know what to let y = .. (they done sin³x, how would you know this?)

For (a) and (b) you are expected to spot.

You also need to spot for part (c). Just realize that is of the form

So for such type of integrals you try to differentiate

Hope it makes sense.

[boredom_personified]

(Original post by sabre2th1)

For c), how would you know what to let y = .. (they done sin³x, how would you know this?)

For c), if it helps, I would have used a different substitution of u=sin(x), so that du=cos(x)dx and you can use integration by substitution to rewrite the integral as integral of 3u²du, obtaining an answer of u³ +C = sin³(x) + C . Ultimately it's the same, but I would find it easier than simply spotting the antiderivative. It would also work for other integrals of the form sin^m(x)cosⁿ(x) with at least one of the m or n odd, by suitably rearranging.

As for a) and b), if you didn't immediately notice that the derivative of the denominator is in the numerator, then you would just have to try a few substitutions, eventually reducing to an a/u du integral (to some constant, a). It would be beneficial to practice spotting this instead of wasting time with random substitutions.

[sabre2th1]

(Original post by raheem94)
For (a) and (b) you are expected to spot.

You also need to spot for part (c). Just realize that is of the form

So for such type of integrals you try to differentiate

Hope it makes sense.

That makes sense! Thanks

[sabre2th1]

(Original post by boredom_personified)
For c), if it helps, I would have used a different substitution of u=sin(x), so that du=cos(x)dx and you can use integration by substitution to rewrite the integral as integral of 3u²du, obtaining an answer of u³ +C = sin³(x) + C . Ultimately it's the same, but I would find it easier than simply spotting the antiderivative. It would also work for other integrals of the form sin^m(x)cosⁿ(x) with at least one of the m or n odd, by suitably rearranging.

As for a) and b), if you didn't immediately notice that the derivative of the denominator is in the numerator, then you would just have to try a few substitutions, eventually reducing to an a/u du integral (to some constant, a). It would be beneficial to practice spotting this instead of wasting time with random substitutions.

I haven't got to that section of the chapter, but when I do, I will be sure to look at this post for help! Thanks