Question

Find integral

Answer 1

\[\int\limits_{a}^{b}\sqrt{x}\sin^2(x ^{3/2}-1)\]

Answer 2

let u= x^[3/2]
du=3/2(x^[1/2] dx

Answer 3

better let x^(3/2) -1 =u

altough its the same thing..

Answer 4

Woops. I forgot about that. I did let u=x^3/2 - 1

Answer 5

\[3/2\int\limits_{a}^{b}\sin^2(u) du\]
\[3/2 * \frac{ -\cos^3 u}{ 3} + c\]

Answer 6

Final answer
\[\frac{ -\cos^3(x ^{3/2}-1) }{ 2}+c\]

Answer 7

Can someone confirm this is somewhat correct?

Answer 8

That looks right

Answer 9

no its not right

Answer 10

\[3/2\int\limits_{a}^{b }sinu ^{2}du\]

using cos2u = 1-sin^2

\[3/2\int\limits_{a}^{b}(1-\cos2u)/2 \]

using integral cos2u =\[\frac{ 1 }{ 2 }\sin2u\]

\[\frac{ 3 }{ 4 } ( u-\frac{ 1 }{ 2 } \sin2u)\]

now put the value of u then uper and lower value of integral

Answer 11

@shubhamsrg Would you continue helping me?

Answer 12

integral of sin^2 u is not cos^3 u /3

Answer 13

I would like to make one minor highlight and correction of something early in your work.

\(\displaystyle \int_{a}^{b} \sqrt{x} sin^2 (x^{3/2} - 1) \; \text{d}x\)
We made the substitution: \(u = x^{3/2} - 1\); differentiating both sides, we obtain the differential relation: \(\displaystyle \text{d}u = \frac{3}{2} x^{1/2} \; \text{d}x\).
We want to replace \(\sqrt{x} \; \text{d}x\) in our initial integration expression, so we need to divide both sides of the latter equation by \(\displaystyle \frac{3}{2}\) to make the correct identity: \(\displaystyle \frac{2}{3} \; \text{d}u = \sqrt{x} \; \text{d}x \).
This substitutes back into our initial integration. Note that with u-substitutions, our domain of integration will be different; however, for the sake of simply reversing our substitution, we will omit it and use a placeholder domain \(D\).

\(\displaystyle \int_{a}^{b} \color{#aa0000}{\sqrt{x}} \sin^2 (\color{#0000aa}{x^{3/2} - 1}) \; \color{#aa0000}{\text{d}x}\)
\(\displaystyle \int_{D} \sin^2 \color{#0000aa}{u} \color{#aa0000}{\frac{2}{3} \; \text{d}u}\)
\(\displaystyle \frac{2}{3} \int_{D} \sin^2 u \; \text{d}u\)

In addition, the correct identity of \(\cos 2u\) is: \(\cos 2u = 1 - \color{#22dd00}{2} \sin^2 u\). The difference is that we missed the 2 here earlier.