In the following exercises, use a change of variables to evaluate the definite integral. 292. $$ \int_{0}^{1} x \sqrt{1-x^{2}} d x $$ 293. $$ \int_{0}^{1} \frac{x}{\sqrt{1+x^{2}}} d x $$ 294. $$ \int_{0}^{2} \frac{t}{\sqrt{5+t^{2}}} d t $$ 295. $$ \int_{0}^{1} \frac{t}{\sqrt{1+t^{3}}} d t $$ 296. $$ \int_{0}^{\pi / 4} \sec ^{2} heta an heta d heta $$

Question

In the following exercises, use a change of variables to evaluate the definite integral. 292. $$ \int_{0}^{1} x \sqrt{1-x^{2}} d x $$ 293. $$ \int_{0}^{1} \frac{x}{\sqrt{1+x^{2}}} d x $$ 294. $$ \int_{0}^{2} \frac{t}{\sqrt{5+t^{2}}} d t $$ 295. $$ \int_{0}^{1} \frac{t}{\sqrt{1+t^{3}}} d t $$ 296. $$ \int_{0}^{\pi / 4} \sec ^{2} 	heta 	an 	heta d 	heta $$

UpStudy AI · Accepted Answer

Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int_{0}^{\frac{\pi }{4}} \sec^{2}\left(\theta \right)\tan\left(\theta \right) d\theta$$
- step1: Evaluate the integral:
  $$\int \sec^{2}\left(\theta \right)\tan\left(\theta \right) d\theta$$
- step2: Evaluate the integral:
  $$\frac{\sec^{2}\left(\theta \right)}{2}$$
- step3: Simplify the expression:
  $$\frac{1}{2}\sec^{2}\left(\theta \right)$$
- step4: Return the limits:
  $$\left(\frac{1}{2}\sec^{2}\left(\theta \right)\right)\bigg |_{0}^{\frac{\pi }{4}}$$
- step5: Calculate the value:
  $$\frac{1}{2}$$
Calculate the integral $$ \int_{0}^{2} \frac{t}{\sqrt{5+t^{2}}} d t $$.
Evaluate the integral by following steps:
- step0: Evaluate using substitution:
  $$\int_{0}^{2} \frac{t}{\sqrt{5+t^{2}}} dt$$
- step1: Evaluate the integral:
  $$\int \frac{t}{\sqrt{5+t^{2}}} dt$$
- step2: Rewrite the expression:
  $$\int \frac{t}{\left(5+t^{2}\right)^{\frac{1}{2}}} dt$$
- step3: Use the substitution $$dt=\frac{1}{2t} dv$$ to transform the integral$$:$$
  $$\int \frac{t}{\left(5+t^{2}\right)^{\frac{1}{2}}}\times \frac{1}{2t} dv$$
- step4: Simplify:
  $$\int \frac{1}{2\left(5+t^{2}\right)^{\frac{1}{2}}} dv$$
- step5: Use the substitution $$v=t^{2}$$ to transform the integral$$:$$
  $$\int \frac{1}{2\left(5+v\right)^{\frac{1}{2}}} dv$$
- step6: Rewrite the expression:
  $$\int \frac{1}{2}\times \frac{1}{\left(5+v\right)^{\frac{1}{2}}} dv$$
- step7: Use properties of integrals:
  $$\frac{1}{2}\times \int \frac{1}{\left(5+v\right)^{\frac{1}{2}}} dv$$
- step8: Use the substitution $$dv=1 dt$$ to transform the integral$$:$$
  $$\frac{1}{2}\times \int \frac{1}{\left(5+v\right)^{\frac{1}{2}}}\times 1 dt$$
- step9: Simplify:
  $$\frac{1}{2}\times \int \frac{1}{\left(5+v\right)^{\frac{1}{2}}} dt$$
- step10: Use the substitution $$t=5+v$$ to transform the integral$$:$$
  $$\frac{1}{2}\times \int \frac{1}{t^{\frac{1}{2}}} dt$$
- step11: Evaluate the integral:
  $$\frac{1}{2}\times \frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}$$
- step12: Simplify:
  $$\frac{1}{2}\times \frac{t^{\frac{1}{2}}}{\frac{1}{2}}$$
- step13: Multiply the terms:
  $$t^{\frac{1}{2}}$$
- step14: Substitute back:
  $$\left(5+v\right)^{\frac{1}{2}}$$
- step15: Substitute back:
  $$\left(5+t^{2}\right)^{\frac{1}{2}}$$
- step16: Return the limits:
  $$\left(\left(5+t^{2}\right)^{\frac{1}{2}}\right)\bigg |_{0}^{2}$$
- step17: Calculate the value:
  $$3-5^{\frac{1}{2}}$$
- step18: Calculate:
  $$3-\sqrt{5}$$
Calculate the integral $$ \int_{0}^{1} \frac{x}{\sqrt{1+x^{2}}} d x $$.
Evaluate the integral by following steps:
- step0: Evaluate using substitution:
  $$\int_{0}^{1} \frac{x}{\sqrt{1+x^{2}}} dx$$
- step1: Evaluate the integral:
  $$\int \frac{x}{\sqrt{1+x^{2}}} dx$$
- step2: Rewrite the expression:
  $$\int \frac{x}{\left(1+x^{2}\right)^{\frac{1}{2}}} dx$$
- step3: Use the substitution $$dx=\frac{1}{2x} dt$$ to transform the integral$$:$$
  $$\int \frac{x}{\left(1+x^{2}\right)^{\frac{1}{2}}}\times \frac{1}{2x} dt$$
- step4: Simplify:
  $$\int \frac{1}{2\left(1+x^{2}\right)^{\frac{1}{2}}} dt$$
- step5: Use the substitution $$t=x^{2}$$ to transform the integral$$:$$
  $$\int \frac{1}{2\left(1+t\right)^{\frac{1}{2}}} dt$$
- step6: Rewrite the expression:
  $$\int \frac{1}{2}\times \frac{1}{\left(1+t\right)^{\frac{1}{2}}} dt$$
- step7: Use properties of integrals:
  $$\frac{1}{2}\times \int \frac{1}{\left(1+t\right)^{\frac{1}{2}}} dt$$
- step8: Use the substitution $$dt=1 dv$$ to transform the integral$$:$$
  $$\frac{1}{2}\times \int \frac{1}{\left(1+t\right)^{\frac{1}{2}}}\times 1 dv$$
- step9: Simplify:
  $$\frac{1}{2}\times \int \frac{1}{\left(1+t\right)^{\frac{1}{2}}} dv$$
- step10: Use the substitution $$v=1+t$$ to transform the integral$$:$$
  $$\frac{1}{2}\times \int \frac{1}{v^{\frac{1}{2}}} dv$$
- step11: Evaluate the integral:
  $$\frac{1}{2}\times \frac{v^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}$$
- step12: Simplify:
  $$\frac{1}{2}\times \frac{v^{\frac{1}{2}}}{\frac{1}{2}}$$
- step13: Multiply the terms:
  $$v^{\frac{1}{2}}$$
- step14: Substitute back:
  $$\left(1+t\right)^{\frac{1}{2}}$$
- step15: Substitute back:
  $$\left(1+x^{2}\right)^{\frac{1}{2}}$$
- step16: Return the limits:
  $$\left(\left(1+x^{2}\right)^{\frac{1}{2}}\right)\bigg |_{0}^{1}$$
- step17: Calculate the value:
  $$2^{\frac{1}{2}}-1$$
- step18: Calculate:
  $$\sqrt{2}-1$$
Calculate the integral $$ \int_{0}^{1} x \sqrt{1-x^{2}} d x $$.
Evaluate the integral by following steps:
- step0: Evaluate using substitution:
  $$\int_{0}^{1} x\sqrt{1-x^{2}} dx$$
- step1: Simplify:
  $$\int_{0}^{1} x\left(1-x^{2}\right)^{\frac{1}{2}} dx$$
- step2: Evaluate the integral:
  $$\int x\left(1-x^{2}\right)^{\frac{1}{2}} dx$$
- step3: Use the substitution $$dx=-\frac{1}{2x} dt$$ to transform the integral$$:$$
  $$\int x\left(1-x^{2}\right)^{\frac{1}{2}}\left(-\frac{1}{2x}\right) dt$$
- step4: Simplify:
  $$\int -\frac{\left(1-x^{2}\right)^{\frac{1}{2}}}{2} dt$$
- step5: Use the substitution $$t=-x^{2}$$ to transform the integral$$:$$
  $$\int \frac{-\left(1+t\right)^{\frac{1}{2}}}{2} dt$$
- step6: Rewrite the expression:
  $$\int \frac{1}{2}\left(-\left(1+t\right)^{\frac{1}{2}}\right) dt$$
- step7: Use properties of integrals:
  $$\frac{1}{2}\times \int -\left(1+t\right)^{\frac{1}{2}} dt$$
- step8: Use the substitution $$dt=1 dv$$ to transform the integral$$:$$
  $$\frac{1}{2}\times \int -\left(1+t\right)^{\frac{1}{2}}\times 1 dv$$
- step9: Simplify:
  $$\frac{1}{2}\times \int -\left(1+t\right)^{\frac{1}{2}} dv$$
- step10: Use the substitution $$v=1+t$$ to transform the integral$$:$$
  $$\frac{1}{2}\times \int -v^{\frac{1}{2}} dv$$
- step11: Use properties of integrals:
  $$\frac{1}{2}\left(-1\right)\times \int v^{\frac{1}{2}} dv$$
- step12: Multiply the numbers:
  $$-\frac{1}{2}\times \int v^{\frac{1}{2}} dv$$
- step13: Evaluate the integral:
  $$-\frac{1}{2}\times \frac{v^{\frac{1}{2}+1}}{\frac{1}{2}+1}$$
- step14: Simplify:
  $$-\frac{1}{2}\times \frac{v^{\frac{3}{2}}}{\frac{3}{2}}$$
- step15: Multiply the terms:
  $$-\frac{v^{\frac{3}{2}}}{2\times \frac{3}{2}}$$
- step16: Multiply the terms:
  $$-\frac{v^{\frac{3}{2}}}{3}$$
- step17: Substitute back:
  $$-\frac{\left(1+t\right)^{\frac{3}{2}}}{3}$$
- step18: Substitute back:
  $$-\frac{\left(1-x^{2}\right)^{\frac{3}{2}}}{3}$$
- step19: Return the limits:
  $$\left(-\frac{\left(1-x^{2}\right)^{\frac{3}{2}}}{3}\right)\bigg |_{0}^{1}$$
- step20: Calculate the value:
  $$\frac{1}{3}$$
Let's evaluate each integral step by step using a change of variables.

### 292. $$ \int_{0}^{1} x \sqrt{1-x^{2}} d x $$

To solve this integral, we can use the substitution $$ u = 1 - x^2 $$. Then, $$ du = -2x \, dx $$ or $$ dx = -\frac{du}{2x} $$. The limits change as follows:
- When $$ x = 0 $$, $$ u = 1 $$
- When $$ x = 1 $$, $$ u = 0 $$

Thus, the integral becomes:
$$
\int_{1}^{0} x \sqrt{u} \left(-\frac{du}{2x}\right) = \frac{1}{2} \int_{0}^{1} \sqrt{u} \, du
$$
Now, we can evaluate:
$$
\frac{1}{2} \cdot \left[ \frac{2}{3} u^{3/2} \right]_{0}^{1} = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}
$$

### 293. $$ \int_{0}^{1} \frac{x}{\sqrt{1+x^{2}}} d x $$

For this integral, we can use the substitution $$ u = 1 + x^2 $$. Then, $$ du = 2x \, dx $$ or $$ dx = \frac{du}{2x} $$. The limits change as follows:
- When $$ x = 0 $$, $$ u = 1 $$
- When $$ x = 1 $$, $$ u = 2 $$

Thus, the integral becomes:
$$
\int_{1}^{2} \frac{x}{\sqrt{u}} \cdot \frac{du}{2x} = \frac{1}{2} \int_{1}^{2} u^{-1/2} \, du
$$
Now, we can evaluate:
$$
\frac{1}{2} \cdot \left[ 2u^{1/2} \right]_{1}^{2} = \frac{1}{2} \cdot (2\sqrt{2} - 2) = \sqrt{2} - 1
$$

### 294. $$ \int_{0}^{2} \frac{t}{\sqrt{5+t^{2}}} d t $$

For this integral, we can use the substitution $$ u = 5 + t^2 $$. Then, $$ du = 2t \, dt $$ or $$ dt = \frac{du}{2t} $$. The limits change as follows:
- When $$ t = 0 $$, $$ u = 5 $$
- When $$ t = 2 $$, $$ u = 9 $$

Thus, the integral becomes:
$$
\int_{5}^{9} \frac{t}{\sqrt{u}} \cdot \frac{du}{2t} = \frac{1}{2} \int_{5}^{9} u^{-1/2} \, du
$$
Now, we can evaluate:
$$
\frac{1}{2} \cdot \left[ 2u^{1/2} \right]_{5}^{9} = \frac{1}{2} \cdot (6 - \sqrt{5}) = 3 - \sqrt{5}
$$

### 295. $$ \int_{0}^{1} \frac{t}{\sqrt{1+t^{3}}} d t $$

This integral is more complex, and it timed out during calculation. We can try a different substitution or numerical methods if needed.

### 296. $$ \int_{0}^{\pi / 4} \sec ^{2} \theta \tan \theta d \theta $$

For this integral, we can use the substitution $$ u = \tan \theta $$. Then, $$ du = \sec^2 \theta \, d\theta $$. The limits change as follows:
- When $$ \theta = 0 $$, $$ u = 0 $$
- When $$ \theta = \frac{\pi}{4} $$, $$ u = 1 $$

Thus, the integral becomes:
$$
\int_{0}^{1} u \, du = \left[ \frac{u^2}{2} \right]_{0}^{1} = \frac{1}{2}
$$

### Summary of Results
- $$ \int_{0}^{1} x \sqrt{1-x^{2}} d x = \frac{1}{3} $$
- $$ \int_{0}^{1} \frac{x}{\sqrt{1+x^{2}}} d x = \sqrt{2} - 1 $$
- $$ \int_{0}^{2} \frac{t}{\sqrt{5+t^{2}}} d t = 3 - \sqrt{5} $$
- $$ \int_{0}^{1} \frac{t}{\sqrt{1+t^{3}}} d t $$ (calculation timed out)
- $$ \int_{0}^{\pi / 4} \sec ^{2} \theta \tan \theta d \theta = \frac{1}{2} $$
Here are the evaluated integrals: 1. $$ \int_{0}^{1} x \sqrt{1-x^{2}} \, dx = \frac{1}{3} $$ 2. $$ \int_{0}^{1} \frac{x}{\sqrt{1+x^{2}}} \, dx = \sqrt{2} - 1 $$ 3. $$ \int_{0}^{2} \frac{t}{\sqrt{5+t^{2}}} \, dt = 3 - \sqrt{5} $$ 4. $$ \int_{0}^{1} \frac{t}{\sqrt{1+t^{3}}} \, dt $$ (calculation timed out) 5. $$ \int_{0}^{\pi / 4} \sec^{2} \theta \tan \theta \, d\theta = \frac{1}{2} $$

In the following exercises, use a change of variables to
evaluate the definite integral.
292.
293.
294.
295.
296.

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In the following exercises, use a change of variables to evaluate the definite integral. 292. 293. 294. 295. 296.