- #1

chwala

Gold Member

- 2,692

- 354

- Homework Statement
- This is my own question (refreshing on the fundamental laws of calculus).

Differentiate the following integral:

$$\int_{x^2}^{2x} \sin t \, dt$$

- Relevant Equations
- calculus

My take:

$$\int_{x^2}^{2x} \sin t \, dt$$

using the

$$\int_{x^2}^{2x} \sin t \, dt=-2x \sin x^2 +2 \sin 2x$$

I also wanted to check my answer, i did this by,

$$\int [-2x \sin x^2 +2 \sin 2x] dx$$

for the integration of the first part i.e

$$\int -2x \sin x^2 dx$$

i let ##u=x^2## giving me,

$$\int -\sin u du=\cos u+k = cos (x^2)+k$$and for the integration of the second part i.e

$$\int 2 \sin 2x dx$$

i let ##u=2x##, giving me,

$$\int 2 \sin 2x dx=\int -\sin u du=-\cos u +k=-\cos 2x+k$$

thus,

$$\int -2x \sin x^2 +2 \sin 2x dx=\cos x^2-\cos 2x+k$$

i just checked with wolframalpha and confirmed that:

$$\int_{x^2}^{2x} \sin t \, dt=\cos x^2-\cos 2x$$

I may get more insight or different ways of proof from you guys.

$$\int_{x^2}^{2x} \sin t \, dt$$

using the

**fundamental theorem of calculus**we shall have,$$\int_{x^2}^{2x} \sin t \, dt=-2x \sin x^2 +2 \sin 2x$$

I also wanted to check my answer, i did this by,

$$\int [-2x \sin x^2 +2 \sin 2x] dx$$

for the integration of the first part i.e

$$\int -2x \sin x^2 dx$$

i let ##u=x^2## giving me,

$$\int -\sin u du=\cos u+k = cos (x^2)+k$$and for the integration of the second part i.e

$$\int 2 \sin 2x dx$$

i let ##u=2x##, giving me,

$$\int 2 \sin 2x dx=\int -\sin u du=-\cos u +k=-\cos 2x+k$$

thus,

$$\int -2x \sin x^2 +2 \sin 2x dx=\cos x^2-\cos 2x+k$$

i just checked with wolframalpha and confirmed that:

$$\int_{x^2}^{2x} \sin t \, dt=\cos x^2-\cos 2x$$

I may get more insight or different ways of proof from you guys.

Last edited: