stuck on question 16c.

I don't understand how 1/2 is placed outside the intergral.

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Original post by Vincent404

I don't understand how 1/2 is placed outside the intergral.

image-af9dbbdc-68d3-4655-a74d-abc5d6997af38842446167850912784-compressed.jpg.jpeg

I presume you're refering to the markscheme which you've not posted, so I'd guess:

Via the power series expansion:

\displaystyle\int \sqrt{\cos x}\;dx =\int\sqrt{1-\frac{x^2}{2}+\cdots}\;dx

\displaystyle\approx\int\sqrt{1-\frac{x^2}{2}}\;dx

\displaystyle=\int\frac{1}{2}\sqrt{4-2x^2}\;dx

Pull the half out before the integral, and use the binomial on the integrand.

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