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Sagot :
Sure, let's tackle the definite integrals one by one.
### (i) [tex]\(\int_0^{\frac{\pi}{3}} \tan^3(x) \sec^2(x) \, dx\)[/tex]
To evaluate this integral, we can use the substitution method. Observe that the integrand contains [tex]\(\tan^3(x)\)[/tex] and [tex]\(\sec^2(x)\)[/tex]. Knowing that the derivative of [tex]\(\tan(x)\)[/tex] is [tex]\(\sec^2(x)\)[/tex], we can use the substitution:
[tex]\[ u = \tan(x) \][/tex]
[tex]\[ \frac{du}{dx} = \sec^2(x) \][/tex]
Thus, [tex]\(du = \sec^2(x) \, dx\)[/tex].
Now, we rewrite the integral in terms of [tex]\(u\)[/tex]:
[tex]\[ \int_0^{\frac{\pi}{3}} \tan^3(x) \sec^2(x) \, dx = \int_0^{\tan(\frac{\pi}{3})} u^3 \, du \][/tex]
Evaluating [tex]\(\tan(\frac{\pi}{3})\)[/tex] gives [tex]\( \sqrt{3} \)[/tex]. So, the integral becomes:
[tex]\[ \int_0^{\sqrt{3}} u^3 \, du \][/tex]
We now integrate [tex]\(u^3\)[/tex]:
[tex]\[ \int u^3 \, du = \frac{u^4}{4} \][/tex]
We then evaluate this at the bounds [tex]\(0\)[/tex] and [tex]\(\sqrt{3}\)[/tex]:
[tex]\[ \left[ \frac{u^4}{4} \right]_0^{\sqrt{3}} = \frac{(\sqrt{3})^4}{4} - \frac{0^4}{4} = \frac{9}{4} \][/tex]
Thus, the result of the first integral is:
[tex]\[ \int_0^{\frac{\pi}{3}} \tan^3(x) \sec^2(x) \, dx = \frac{9}{4} = 2.25 \][/tex]
### (ii) [tex]\(\int_0^{\frac{\pi}{4}} \frac{\sec^2(x)}{1 + \tan(x)} \, dx\)[/tex]
For this integral, we can again use substitution. Let's set:
[tex]\[ u = \tan(x) \][/tex]
[tex]\[ \frac{du}{dx} = \sec^2(x) \][/tex]
Thus, [tex]\(du = \sec^2(x) \, dx\)[/tex].
Rewriting the integral in terms of [tex]\(u\)[/tex], we get:
[tex]\[ \int_0^{\frac{\pi}{4}} \frac{\sec^2(x)}{1 + \tan(x)} \, dx = \int_0^{\tan(\frac{\pi}{4})} \frac{du}{1 + u} \][/tex]
Evaluating [tex]\(\tan(\frac{\pi}{4})\)[/tex] gives [tex]\(1\)[/tex]. So, the integral becomes:
[tex]\[ \int_0^{1} \frac{du}{1 + u} \][/tex]
We now integrate [tex]\(\frac{1}{1 + u}\)[/tex]:
[tex]\[ \int \frac{1}{1 + u} \, du = \ln|1 + u| \][/tex]
We then evaluate this at the bounds [tex]\(0\)[/tex] and [tex]\(1\)[/tex]:
[tex]\[ \left[ \ln|1 + u| \right]_0^1 = \ln(1 + 1) - \ln(1 + 0) = \ln(2) - \ln(1) \][/tex]
Since [tex]\(\ln(1)\)[/tex] is 0, the result simplifies to:
[tex]\[ \ln(2) \][/tex]
Thus, the result of the second integral is:
[tex]\[ \int_0^{\frac{\pi}{4}} \frac{\sec^2(x)}{1 + \tan(x)} \, dx = \ln(2) \approx 0.6931471805599453 \][/tex]
### Summary
(i) [tex]\(\int_0^{\frac{\pi}{3}} \tan^3(x) \sec^2(x) \, dx = 2.25\)[/tex]
(ii) [tex]\(\int_0^{\frac{\pi}{4}} \frac{\sec^2(x)}{1 + \tan(x)} \, dx = \ln(2) \approx 0.6931471805599453\)[/tex]
### (i) [tex]\(\int_0^{\frac{\pi}{3}} \tan^3(x) \sec^2(x) \, dx\)[/tex]
To evaluate this integral, we can use the substitution method. Observe that the integrand contains [tex]\(\tan^3(x)\)[/tex] and [tex]\(\sec^2(x)\)[/tex]. Knowing that the derivative of [tex]\(\tan(x)\)[/tex] is [tex]\(\sec^2(x)\)[/tex], we can use the substitution:
[tex]\[ u = \tan(x) \][/tex]
[tex]\[ \frac{du}{dx} = \sec^2(x) \][/tex]
Thus, [tex]\(du = \sec^2(x) \, dx\)[/tex].
Now, we rewrite the integral in terms of [tex]\(u\)[/tex]:
[tex]\[ \int_0^{\frac{\pi}{3}} \tan^3(x) \sec^2(x) \, dx = \int_0^{\tan(\frac{\pi}{3})} u^3 \, du \][/tex]
Evaluating [tex]\(\tan(\frac{\pi}{3})\)[/tex] gives [tex]\( \sqrt{3} \)[/tex]. So, the integral becomes:
[tex]\[ \int_0^{\sqrt{3}} u^3 \, du \][/tex]
We now integrate [tex]\(u^3\)[/tex]:
[tex]\[ \int u^3 \, du = \frac{u^4}{4} \][/tex]
We then evaluate this at the bounds [tex]\(0\)[/tex] and [tex]\(\sqrt{3}\)[/tex]:
[tex]\[ \left[ \frac{u^4}{4} \right]_0^{\sqrt{3}} = \frac{(\sqrt{3})^4}{4} - \frac{0^4}{4} = \frac{9}{4} \][/tex]
Thus, the result of the first integral is:
[tex]\[ \int_0^{\frac{\pi}{3}} \tan^3(x) \sec^2(x) \, dx = \frac{9}{4} = 2.25 \][/tex]
### (ii) [tex]\(\int_0^{\frac{\pi}{4}} \frac{\sec^2(x)}{1 + \tan(x)} \, dx\)[/tex]
For this integral, we can again use substitution. Let's set:
[tex]\[ u = \tan(x) \][/tex]
[tex]\[ \frac{du}{dx} = \sec^2(x) \][/tex]
Thus, [tex]\(du = \sec^2(x) \, dx\)[/tex].
Rewriting the integral in terms of [tex]\(u\)[/tex], we get:
[tex]\[ \int_0^{\frac{\pi}{4}} \frac{\sec^2(x)}{1 + \tan(x)} \, dx = \int_0^{\tan(\frac{\pi}{4})} \frac{du}{1 + u} \][/tex]
Evaluating [tex]\(\tan(\frac{\pi}{4})\)[/tex] gives [tex]\(1\)[/tex]. So, the integral becomes:
[tex]\[ \int_0^{1} \frac{du}{1 + u} \][/tex]
We now integrate [tex]\(\frac{1}{1 + u}\)[/tex]:
[tex]\[ \int \frac{1}{1 + u} \, du = \ln|1 + u| \][/tex]
We then evaluate this at the bounds [tex]\(0\)[/tex] and [tex]\(1\)[/tex]:
[tex]\[ \left[ \ln|1 + u| \right]_0^1 = \ln(1 + 1) - \ln(1 + 0) = \ln(2) - \ln(1) \][/tex]
Since [tex]\(\ln(1)\)[/tex] is 0, the result simplifies to:
[tex]\[ \ln(2) \][/tex]
Thus, the result of the second integral is:
[tex]\[ \int_0^{\frac{\pi}{4}} \frac{\sec^2(x)}{1 + \tan(x)} \, dx = \ln(2) \approx 0.6931471805599453 \][/tex]
### Summary
(i) [tex]\(\int_0^{\frac{\pi}{3}} \tan^3(x) \sec^2(x) \, dx = 2.25\)[/tex]
(ii) [tex]\(\int_0^{\frac{\pi}{4}} \frac{\sec^2(x)}{1 + \tan(x)} \, dx = \ln(2) \approx 0.6931471805599453\)[/tex]
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