Question

Evaluate the following integrals: (i) \(\int_{-1}^{2} \mathrm{e}^{\\{3 x\\}} \mathrm{dx}\) (ii) \(\int_{0}^{41 \pi / 2} \sin x d x\) (iii) \(\int_{\pi}^{7 \pi / 2}|\cos x| d x\) (iv) \(\int_{\pi}^{5 \pi / 4} \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} d x\)

Short answer

Answer: The value of the integral \(\int_{\pi}^{5 \pi / 4} \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} d x\) cannot be found analytically. It requires the use of numerical methods such as Simpson's rule or the trapezoidal rule to approximate its value.

Step-by-step solution

Find the antiderivative of the function

The antiderivative of \(e^{3x}\) is \(\frac{1}{3}e^{3x}\).

Apply the Fundamental Theorem of Calculus

Using the Fundamental Theorem of Calculus, we evaluate the integral as follows: \(\int_{-1}^{2} \mathrm{e}^{\\{3 x\\}} \mathrm{dx} = \left[\frac{1}{3}e^{3x}\right]_{-1}^{2} = \frac{1}{3}(e^{6} - e^{-3})\) The value of the integral is \(\frac{1}{3}(e^{6} - e^{-3})\). (ii) \(\int_{0}^{41 \pi / 2} \sin x d x\)

Find the antiderivative of the function

The antiderivative of \(\sin x\) is \(-\cos x\).

Apply the Fundamental Theorem of Calculus

Using the Fundamental Theorem of Calculus, we evaluate the integral as follows: \(\int_{0}^{41 \pi / 2} \sin x d x = \left[-\cos x\right]_{0}^{41 \pi / 2} = -\cos(41\pi / 2) + \cos(0)\) Note that since \(\cos(41\pi / 2) = 1\) and \(\cos(0) = 1\), the value of the integral is \(-1 + 1 = 0\). (iii) \(\int_{\pi}^{7 \pi / 2}|\cos x| d x\)

Find the antiderivative of the function

The antiderivative of \(|\cos x|\) is \(x + \sin x\) when \(\cos x \ge 0\) and \(x - \sin x\) when \(\cos x < 0\).

Evaluate the integral in intervals

Since the absolute value function is not continuous, we will break the integral into intervals where \(\cos x \ge 0\) and \(\cos x < 0\): For \(\pi \le x < 3\pi/2\), we have \(\cos x < 0\). Therefore, the integral is \(\int_{\pi}^{3\pi/2} (x - \sin x) d x\). For \(3\pi/2 \le x \le 7 \pi / 2\), we have \(\cos x \geq 0\). Therefore, the integral is \(\int_{3\pi/2}^{7\pi/2} (x + \sin x) d x\). Now, evaluate the integrals: \(\int_{\pi}^{3\pi/2} (x - \sin x) d x = \left[\frac{x^2}{2} + \cos x\right]_{\pi}^{3 \pi / 2} = -\frac{9\pi^2}{4} + \pi^2 = \frac{-5\pi^2}{4}\) \(\int_{3\pi/2}^{7\pi/2} (x + \sin x) d x = \left[\frac{x^2}{2} - \cos x\right]_{3 \pi / 2}^{7 \pi / 2} = \frac{49\pi^2}{4} - \frac{9\pi^2}{4} = 10\pi^2\) Summing the integrals within the intervals, we get: \(\int_{\pi}^{7 \pi / 2}|\cos x| d x=\frac{-5\pi^2}{4} + 10\pi^2 = \frac{35\pi^2}{4}\) (iv) \(\int_{\pi}^{5 \pi / 4} \frac{\sin 2 x}{\sin ^{4} x+\cos ^{4} x} d x\)

Simplify the function

Use the trigonometric identity \(\sin 2x = 2\sin x \cos x\) and rewrite the integral as: \(\int_{\pi}^{5 \pi / 4} \frac{2\sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x\)

Perform the substitution

Let \(u = \sin x\). Then, \(\mathrm{d}u = \cos x \mathrm{d}x\). Now rewrite the integral with the substitution: \(\int_{\pi}^{5 \pi / 4} \frac{2\sin x \cos x}{\sin ^{4} x+\cos ^{4} x} dx = \int_{\sin \pi}^{\sin (5\pi / 4)} \frac{2u}{u^4 + (1 - u^2)^2} du = \int_{0}^{1/\sqrt{2}} \frac{2u}{u^4 + (1 - u^2)^2} du\) We can then make use of the symmetry around the y-axis to rewrite the integral as follows: \(2 \int_{0}^{1/\sqrt{2}} \frac{u}{u^4 + (1 - u^2)^2} du\) However, solving this integral analytically is very difficult. This problem can be tackled using numerical methods such as Simpson's rule or trapezoidal rule to approximate the value of the integral.