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Exercise 7.8 · Q17

Q.Evaluate the definite integral: ∫0π/4(2sec⁡2x+x3+2) dx\int_{0}^{\pi/4} (2 \sec^2 x + x^3 + 2) \, dx

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The integral splits into three simpler terms. The sec⁡2x\sec^2 x term gives tan⁡x\tan x, the polynomial terms integrate directly, and evaluating from 00 to π/4\pi/4 yields 2+π41024+π22 + \frac{\pi^4}{1024} + \frac{\pi}{2}.

We are asked to evaluate

∫0π/4(2sec⁡2x+x3+2) dx.\int_{0}^{\pi/4} (2 \sec^2 x + x^3 + 2) \, dx.

The key idea is that the integral of a sum is the sum of the integrals. Each term here has a straightforward antiderivative — no symmetry tricks needed, just direct integration. Let’s go term by term.

  1. Integrate 2sec⁡2x2\sec^2 x Recall that ddx(tan⁡x)=sec⁡2x\frac{d}{dx}(\tan x) = \sec^2 x. So

∫2sec⁡2x dx=2tan⁡x+C.\int 2\sec^2 x \, dx = 2\tan x + C.

  1. Integrate x3x^3 Using the power rule ∫xndx=xn+1n+1\int x^n dx = \frac{x^{n+1}}{n+1} for n≠−1n \neq -1:

∫x3 dx=x44+C.\int x^3 \, dx = \frac{x^4}{4} + C.

  1. Integrate the constant 22

∫2 dx=2x+C.\int 2 \, dx = 2x + C.

Now combine these antiderivatives. An antiderivative of the whole integrand is

F(x)=2tan⁡x+x44+2x.F(x) = 2\tan x + \frac{x^4}{4} + 2x.

  1. Evaluate from 00 to π/4\pi/4 By the Fundamental Theorem of Calculus:

∫0π/4(2sec⁡2x+x3+2) dx=F ⁣(π4)−F(0).\int_{0}^{\pi/4} (2 \sec^2 x + x^3 + 2) \, dx = F\!\left(\frac{\pi}{4}\right) - F(0).

Compute F(π/4)F(\pi/4):

tan⁡(π4)=1,so 2tan⁡(π4)=2.\tan\left(\frac{\pi}{4}\right) = 1, \quad \text{so } 2\tan\left(\frac{\pi}{4}\right) = 2.

(π/4)44=π444⋅4=π4256⋅4=π41024.\frac{(\pi/4)^4}{4} = \frac{\pi^4}{4^4 \cdot 4} = \frac{\pi^4}{256 \cdot 4} = \frac{\pi^4}{1024}.

2⋅π4=π2.2 \cdot \frac{\pi}{4} = \frac{\pi}{2}.

Hence

F ⁣(π4)=2+π41024+π2.F\!\left(\frac{\pi}{4}\right) = 2 + \frac{\pi^4}{1024} + \frac{\pi}{2}.

Compute F(0)F(0): …

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