Find the indefinite integral ∫xsin⁡x2dx

Question

Phisecome · Accepted Answer

Step 1  Given: The integral ∫xsin⁡x2dx,  To evaluate: The given indefinite integral.  Step 2  Explanation:  Let I=∫xsin⁡x2dx  Substituting   x2=t  ⇒2xdx=dt  ⇒xdx=dt2  Hence  I=∫sin⁡tdt2  ⇒I=12∫sin⁡tdt  ⇒I=−12cos⁡t+C   [∵∫sin⁡xdx=−cos⁡x]  Now substituting t=x2,  I=−cos⁡x22+C  Where C is arbitrary constant known as constant of integration.  Answer: ∫xsin⁡x2dx=−cos⁡x22+C, where C is constant of integration.

Befory · Accepted Answer

Step 1: Use Integration by Substitution.
Let

u = x^{2}, d u = 2 x d x, t h e n x d x = \frac{1}{2} d u

Step 2: Using u and du above, rewrite

\int x \sin (x^{2}) d x

.

\int \frac{\sin u}{2} d u

Step 3: Use Constant Factor Rule:

\int c f (x) d x = c \int f (x) d x

.

\frac{1}{2} \int \sin u d u

Step 4: Use Trigonometric Integration: the integral of

\sin u i s - \cos u

.

- \frac{\cos u}{2}

Step 5: Substitute

u = x^{2}

back into the original integral.

- \frac{\cos (x^{2})}{2}

Step 6: Add constant.

- \frac{\cos (x^{2})}{2} + C

Find the indefinite integral \int x\sin x^{2}dx

Answered question

Answer & Explanation

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