Q.Evaluate the following integrals using properties of integration:

Step 1 (i). Determine the parity of $g(x)=\cos\!\left(\dfrac{e^x-1}{e^x+1}\right)$.

Multiply numerator and denominator by $e^x$:

So $g(-x)=\cos\!\left(-\dfrac{e^x-1}{e^x+1}\right)=\cos\!\left(\dfrac{e^x-1}{e^x+1}\right)=g(x)$ (cosine is even) — $g$ is even.

Step 2 (i). Determine the parity of the full integrand and apply Property 9.

The integrand is $f(x)=x\,g(x)$. Since $x$ is odd and $g$ is even, $f(-x)=(-x)g(-x)=-x\,g(x)=-f(x)$, so $f$ is odd. By Property 9 (odd-function shortcut), $\displaystyle\int_{-a}^{a}f(x)\,dx=0$ for any odd $f$. Hence

Step 3 (ii). Split the integral by linearity (Property 4) and classify each term's parity on $[-\pi/2,\pi/2]$.

• $x^5$: $(-x)^5=-x^5$, odd.
• $x\cos x$: $x$ is odd, $\cos x$ is even, so the product is odd: $(-x)\cos(-x)=-x\cos x$.
• $\tan^3x$: $\tan(-x)=-\tan x\Rightarrow\tan^3(-x)=-\tan^3x$, odd.
• $1$: even (constant).

Step 4 (ii). Apply Property 9 to the three odd terms and evaluate the constant term.

By Property 9 the first three integrals all vanish. For the last, $\displaystyle\int_{-\pi/2}^{\pi/2}1\,dx=\pi/2-(-\pi/2)=\pi$. So

Step 5 (iii). Use the even-function shortcut (Property 8).

$\sin^2x$ is even, since $\sin^2(-x)=(-\sin x)^2=\sin^2x$. So

Step 6 (iii). Evaluate with the half-angle formula.

$\sin^2x=\dfrac{1-\cos2x}{2}$, so

Doubling: $\displaystyle\int_{-\pi/4}^{\pi/4}\sin^2x\,dx=2\left(\frac{\pi}{8}-\frac14\right)=\frac{\pi}{4}-\frac12.$

Step 7 (iv). Reflect the original integral $I=\displaystyle\int_0^{2\pi}x\log\!\left(\dfrac{3+\cos x}{3-\cos x}\right)dx$ using $x\to2\pi-x$ (Property 6).

Write $h(x)=\log\!\left(\dfrac{3+\cos x}{3-\cos x}\right)$. Since $\cos(2\pi-x)=\cos x$ (period $2\pi$), $h(2\pi-x)=h(x)$. By Property 6 (with $a=0,\ b=2\pi$),

Step 8 (iv). Solve for $I$ in terms of $J=\int_0^{2\pi}h(x)\,dx$, then show $J=0$.

$2I=2\pi J\Rightarrow I=\pi J$. Split $J$ at $\pi$ (Property 3, additivity) and shift the second piece by $x=t+\pi$:

Since $\cos(t+\pi)=-\cos t$, $h(t+\pi)=\log\!\left(\dfrac{3-\cos t}{3+\cos t}\right)=-\log\!\left(\dfrac{3+\cos t}{3-\cos t}\right)=-h(t)$. So

Hence $I=\pi\cdot0=0$.

Step 9 (v). Rewrite $\cos^3x$ using $\cos^2x=1-\sin^2x$, then split by linearity (Property 4).

so

Step 10 (v). Evaluate each piece as a full-period total-derivative integral.

$\sin^4x\cos x=\dfrac{d}{dx}\!\left(\dfrac{\sin^5x}{5}\right)$, so $\displaystyle\int_0^{2\pi}\sin^4x\cos x\,dx=\left[\frac{\sin^5x}{5}\right]_0^{2\pi}=\frac{\sin^5(2\pi)-\sin^5(0)}{5}=0$ (since $\sin0=\sin2\pi=0$). Likewise $\sin^6x\cos x=\dfrac{d}{dx}\!\left(\dfrac{\sin^7x}{7}\right)$, giving $\displaystyle\int_0^{2\pi}\sin^6x\cos x\,dx=\left[\frac{\sin^7x}{7}\right]_0^{2\pi}=0$. So

Step 11 (vi). Locate the sign-change point of $5x-3$ on $[0,1]$.

$5x-3=0$ at $x=\tfrac35\in[0,1]$; $5x-3<0$ on $[0,\tfrac35)$ and $5x-3>0$ on $(\tfrac35,1]$. By additivity (Property 3),

Step 12 (vi). Evaluate both pieces.

Adding: $\displaystyle\int_0^1|5x-3|\,dx=\frac{9}{10}+\frac25=\frac{13}{10}$ (this is exactly the sum of the two triangular areas of the $V$-shaped graph — base $\tfrac35$, height $3$, plus base $\tfrac25$, height $2$).

Step 13 (vii). Differentiate $F(x)=\displaystyle\int_0^{\sin^2x}\sin^{-1}\sqrt t\,dt+\int_0^{\cos^2x}\cos^{-1}\sqrt t\,dt$ using the First Fundamental Theorem plus the chain rule (on $0\le x\le\pi/2$, where $\sin x,\cos x\ge0$).

On $[0,\pi/2]$, $\sin^{-1}(\sin x)=x$ and $\cos^{-1}(\cos x)=x$, so

Step 14 (vii). $F$ is constant — evaluate it at the convenient point $x=0$.

Since $F'(x)\equiv0$ on $[0,\pi/2]$, $F(x)=F(0)$ for every $x$ in this interval. At $x=0$: $\sin^2(0)=0$ and $\cos^2(0)=1$, so

Step 15 (vii). Evaluate $\displaystyle\int_0^1\cos^{-1}\sqrt t\,dt$ by the substitution $t=u^2$.

$dt=2u\,du$, $u:0\to1$:

Integrate by parts ($p=\cos^{-1}u,\ dq=u\,du$): $\displaystyle\int_0^1u\cos^{-1}u\,du=\left[\frac{u^2}{2}\cos^{-1}u\right]_0^1+\frac12\int_0^1\frac{u^2}{\sqrt{1-u^2}}\,du=0+\frac12\int_0^1\frac{u^2}{\sqrt{1-u^2}}\,du$ (the boundary term vanishes since $\cos^{-1}1=0$ and the $u=0$ term is $0$). Substituting $u=\sin\theta$: $\displaystyle\int_0^1\frac{u^2}{\sqrt{1-u^2}}\,du=\int_0^{\pi/2}\sin^2\theta\,d\theta=\frac{\pi}{4}$. So $\displaystyle\int_0^1u\cos^{-1}u\,du=\frac{\pi}{8}$, and $\displaystyle\int_0^1\cos^{-1}\sqrt t\,dt=2\cdot\frac{\pi}{8}=\frac{\pi}{4}$. Hence $F(x)\equiv\dfrac{\pi}{4}$ for every $x\in[0,\pi/2]$ (self-check: repeating the identical computation for $\int_0^1\sin^{-1}\sqrt t\,dt$ gives $\pi/4$ too, so $F(\pi/2)=\pi/4$ as well, confirming the constant is consistent at both ends).

Step 16 (viii). Convert to a trig integral via $x=\tan\theta$.

$dx=\sec^2\theta\,d\theta$, $1+x^2=\sec^2\theta$; $x:0\to1\Rightarrow\theta:0\to\pi/4$. So

Step 17 (viii). Reflect with $\theta\to\pi/4-\theta$ (Property 6) and add.

$\tan\!\left(\dfrac{\pi}{4}-\theta\right)=\dfrac{1-\tan\theta}{1+\tan\theta}$, so $1+\tan\!\left(\dfrac{\pi}{4}-\theta\right)=\dfrac{(1+\tan\theta)+(1-\tan\theta)}{1+\tan\theta}=\dfrac{2}{1+\tan\theta}$. Hence

So $2I=\dfrac{\pi}{4}\log2\Rightarrow I=\dfrac{\pi\log2}{8}$.

Step 18 (ix). Reflect $I=\displaystyle\int_0^\pi\frac{x\sin x}{1+\sin x}\,dx$ with $x\to\pi-x$ (Property 6) and add.

$\sin(\pi-x)=\sin x$, so

Step 19 (ix). Evaluate $K=\displaystyle\int_0^\pi\frac{\sin x}{1+\sin x}\,dx=\int_0^\pi\left[1-\frac{1}{1+\sin x}\right]dx=\pi-L$, where $L=\displaystyle\int_0^\pi\frac{dx}{1+\sin x}$.

Use the Weierstrass substitution $t=\tan(x/2)$: $\sin x=\dfrac{2t}{1+t^2}$, $dx=\dfrac{2\,dt}{1+t^2}$, $x:0\to\pi\Rightarrow t:0\to\infty$ (an improper integral, per the Improper Integrals topic). …

Properties of definite integrals — parity (even/odd) shortcuts (Props 8-9), the $a+b-x$ reflect-and-add trick (Prop 6), half-period doubling/cancellation (Props 10-11), and the $x\,f(x)$ symmetry trick (Prop 12); part (vii) instead differentiates under the integral …

• Missing that $x\cos x$ (part ii) is odd because $x$ is odd and $\cos x$ is even (odd$\times$even = odd) — wrongly treating the whole product as even.
• In part (iv)/(ix)/(x), reflecting the integrand alone and forgetting the extra factor of $x$ also transforms to $(a+b-x)$, which is what produces the solvable $2I=\ldots$ equation.
• In part (xi), assuming $\cos^2(\cos(\pi-x))\ne\cos^2(\cos x)$ because $\cos(\pi-x)=-\cos x$ looks different — forgetting that squaring an EVEN function of the already-negated argument restores equality.
• Quoting the reduction-formula closed form for $\int_0^{\pi/2}\sin^m\cos^n$ (part v/vi style) without checking that the limits here are $0$ to $2\pi$, not $0$ to $\pi/2$, so that formula does not directly apply. …