# Calculating $\int_{0}^{\frac{\pi}{6}} \cos{(x)} \sqrt{2\sin (x)+1} dx$

I am trying to calculate the value of the following $$\int_{0}^{\frac{\pi}{6}} cosx \sqrt{2sin x+1} dx$$

I used a substitution method.

$$u = 2 \sin (x) + 1$$ $$\frac{du}{dx} = 2\cos (x)$$ $$\frac{u}{2 \cos (x)}du = dx$$

hence $$\int_{0}^{\frac{\pi}{6}} \cos (x) \sqrt{2 \sin (x)+1} dx$$ = $$\int_{1}^{2} \cos (x) \sqrt{u} \times\frac{u}{2 \cos (x)}du$$ = $$\int_{1}^{2} \frac{1}{2}u^\frac{3}{2}du$$ = $$\left[\frac{1}{5}u^\frac{5}{2}\right]_1^2$$

but I can't seem to move any further.

Many thanks.

UPDATE

The third line is incorrect. It should be

$$\frac{du}{2 \cos (x)} = dx$$ hence = $$\int_{1}^{2} \frac{1}{2}u^\frac{1}{2}du$$

= $$\left[\frac{1}{3}u^\frac{3}{2}\right]_1^2$$ = $$\frac{1}{3}\times (2\sqrt3 - 1)$$ = $$\frac{2\sqrt3-1}{3}$$

• Well, didn't you actually solve the whole thing? The answer would be$\frac15(2^{\frac52}-1)$. Apr 22, 2017 at 11:55
• See the updated - the answer is given to be $\frac{2\sqrt2 -1}{3}$ @flytothesurface Apr 22, 2017 at 11:56
• What did you do in your third line of calculations? You seem to have taken a factor $\;u\;$ that wasn't there before...and shouldn't be. Apr 22, 2017 at 12:02
• @DonAntonio seen it yes - so it will be = $\int_{1}^{2} \frac{1}{2}u^\frac{1}{2}du$ hence giving the correct answer. Apr 22, 2017 at 12:06
• @BobSmith Indeed so. A very small yet significative mistake. Apr 22, 2017 at 12:07

$$\frac{du}{dx}=2\cos(x)$$

You should've moved the $dx$ to the other side to get

$$du=2\cos(x)\ dx$$

but you had an extra $u$. Follow this, and the rest of your work is fine.

• Thank you very much. Please upvote if you find it useful for other people to look at, and have a nice day! Apr 22, 2017 at 12:14

$$\int \cos(x) \sqrt{2\sin(x)+1} dx = \frac{1}{2}\int \sqrt{2\sin(x)+1}d(2\sin(x)+1) = \frac{1}{2}(\frac{2}{3}(2\sin(x)+1)\sqrt{2\sin(x)+1}+C)=F(x)$$

So the result is $F(\frac{\pi}{6}) - F(0)$