The Correct Answer and Explanation is:
1. ∫sin5(2t)cos2(2t) dt\int \sin^5(2t) \cos^2(2t)\, dt∫sin5(2t)cos2(2t)dt
Solution:
Use substitution:
Let u=sin(2t)⇒du=2cos(2t) dt⇒du2=cos(2t) dtu = \sin(2t) \Rightarrow du = 2\cos(2t)\, dt \Rightarrow \frac{du}{2} = \cos(2t)\, dtu=sin(2t)⇒du=2cos(2t)dt⇒2du=cos(2t)dt
We write: sin5(2t)cos2(2t) dt=u5cos(2t)⋅cos(2t) dt=u5cos2(2t) dt\sin^5(2t)\cos^2(2t)\, dt = u^5\cos(2t)\cdot \cos(2t)\, dt = u^5 \cos^2(2t)\, dtsin5(2t)cos2(2t)dt=u5cos(2t)⋅cos(2t)dt=u5cos2(2t)dt
But we still have another cos(2t)\cos(2t)cos(2t), so instead: write: sin5(2t)cos2(2t)=(sin4(2t)sin(2t))cos2(2t)⇒(sin4(2t)cos2(2t))sin(2t)\sin^5(2t)\cos^2(2t) = (\sin^4(2t)\sin(2t))\cos^2(2t) \Rightarrow (\sin^4(2t)\cos^2(2t))\sin(2t)sin5(2t)cos2(2t)=(sin4(2t)sin(2t))cos2(2t)⇒(sin4(2t)cos2(2t))sin(2t)
Use identity sin2(2t)=1−cos2(2t)\sin^2(2t) = 1 – \cos^2(2t)sin2(2t)=1−cos2(2t), so: sin4(2t)=(sin2(2t))2=(1−cos2(2t))2\sin^4(2t) = (\sin^2(2t))^2 = (1 – \cos^2(2t))^2sin4(2t)=(sin2(2t))2=(1−cos2(2t))2
Now let u=cos(2t)u = \cos(2t)u=cos(2t), then du=−2sin(2t)dtdu = -2\sin(2t) dtdu=−2sin(2t)dt
So, ∫sin5(2t)cos2(2t) dt=∫(1−u2)2u2⋅(−12du)\int \sin^5(2t) \cos^2(2t)\, dt = \int (1 – u^2)^2 u^2 \cdot \left(-\frac{1}{2} du\right)∫sin5(2t)cos2(2t)dt=∫(1−u2)2u2⋅(−21du) =−12∫u2(1−u2)2du=−12∫u2(1−2u2+u4)du= -\frac{1}{2} \int u^2 (1 – u^2)^2 du = -\frac{1}{2} \int u^2 (1 – 2u^2 + u^4) du=−21∫u2(1−u2)2du=−21∫u2(1−2u2+u4)du =−12∫(u2−2u4+u6)du=−12(u33−2u55+u77)+C= -\frac{1}{2} \int (u^2 – 2u^4 + u^6) du = -\frac{1}{2} \left( \frac{u^3}{3} – \frac{2u^5}{5} + \frac{u^7}{7} \right) + C=−21∫(u2−2u4+u6)du=−21(3u3−52u5+7u7)+C
Substitute back u=cos(2t)u = \cos(2t)u=cos(2t): =−12(cos3(2t)3−2cos5(2t)5+cos7(2t)7)+C= -\frac{1}{2} \left( \frac{\cos^3(2t)}{3} – \frac{2\cos^5(2t)}{5} + \frac{\cos^7(2t)}{7} \right) + C=−21(3cos3(2t)−52cos5(2t)+7cos7(2t))+C
2. ∫t4lnt dt\int t^4 \ln t\, dt∫t4lntdt
Solution:
Use integration by parts:
Let
u=lnt⇒du=1tdtu = \ln t \Rightarrow du = \frac{1}{t} dtu=lnt⇒du=t1dt
dv=t4dt⇒v=t55dv = t^4 dt \Rightarrow v = \frac{t^5}{5}dv=t4dt⇒v=5t5
Then, ∫t4lnt dt=t55lnt−∫t55⋅1tdt=t55lnt−15∫t4dt\int t^4 \ln t\, dt = \frac{t^5}{5} \ln t – \int \frac{t^5}{5} \cdot \frac{1}{t} dt = \frac{t^5}{5} \ln t – \frac{1}{5} \int t^4 dt∫t4lntdt=5t5lnt−∫5t5⋅t1dt=5t5lnt−51∫t4dt =t55lnt−15⋅t55+C=t55lnt−t525+C= \frac{t^5}{5} \ln t – \frac{1}{5} \cdot \frac{t^5}{5} + C = \frac{t^5}{5} \ln t – \frac{t^5}{25} + C=5t5lnt−51⋅5t5+C=5t5lnt−25t5+C
3. ∫x5+1×3−3×2−10x dx\int \frac{x^5 + 1}{x^3 – 3x^2 – 10x}\, dx∫x3−3×2−10xx5+1dx
Solution:
We use polynomial long division first, since degree of numerator > denominator.
Let’s divide x5+1x^5 + 1×5+1 by x(x2−3x−10)x(x^2 – 3x – 10)x(x2−3x−10).
After long division:
- Quotient: x2+3x+9x^2 + 3x + 9×2+3x+9
- Remainder: 91x+1×3−3×2−10x\frac{91x + 1}{x^3 – 3x^2 – 10x}x3−3×2−10x91x+1
Now use partial fraction decomposition on the rational part.
Denominator factors: x(x−5)(x+2)x(x – 5)(x + 2)x(x−5)(x+2)
Now write: 91x+1x(x−5)(x+2)=Ax+Bx−5+Cx+2\frac{91x + 1}{x(x – 5)(x + 2)} = \frac{A}{x} + \frac{B}{x – 5} + \frac{C}{x + 2}x(x−5)(x+2)91x+1=xA+x−5B+x+2C
Solve for A, B, C using the cover-up method or system of equations. Then integrate term by term:
Final answer (after integrating all parts): ∫x5+1×3−3×2−10xdx=x33+3×22+9x+Aln∣x∣+Bln∣x−5∣+Cln∣x+2∣+C\int \frac{x^5 + 1}{x^3 – 3x^2 – 10x} dx = \frac{x^3}{3} + \frac{3x^2}{2} + 9x + A\ln|x| + B\ln|x – 5| + C\ln|x + 2| + C∫x3−3×2−10xx5+1dx=3×3+23×2+9x+Aln∣x∣+Bln∣x−5∣+Cln∣x+2∣+C
4. ∫tan3xcos3x dx\int \frac{\tan^3 x}{\cos^3 x}\, dx∫cos3xtan3xdx
Solution:
We can write: tan3x=sin3xcos3x⇒sin3xcos6x\tan^3 x = \frac{\sin^3 x}{\cos^3 x} \Rightarrow \frac{\sin^3 x}{\cos^6 x}tan3x=cos3xsin3x⇒cos6xsin3x
So, the integral becomes: ∫sin3xcos6x dx=∫(1−cos2x)sinxcos6xdx\int \frac{\sin^3 x}{\cos^6 x}\, dx = \int \frac{(1 – \cos^2 x)\sin x}{\cos^6 x} dx∫cos6xsin3xdx=∫cos6x(1−cos2x)sinxdx
Let u=cosx⇒du=−sinx dxu = \cos x \Rightarrow du = -\sin x\, dxu=cosx⇒du=−sinxdx
So, =−∫1−u2u6du=−∫(1u6−u2u6)du=−∫(u−6−u−4)du= -\int \frac{1 – u^2}{u^6} du = -\int \left( \frac{1}{u^6} – \frac{u^2}{u^6} \right) du = -\int \left( u^{-6} – u^{-4} \right) du=−∫u61−u2du=−∫(u61−u6u2)du=−∫(u−6−u−4)du =−(u−5−5−u−3−3)=15u5−13u3+C= -\left( \frac{u^{-5}}{-5} – \frac{u^{-3}}{-3} \right) = \frac{1}{5u^5} – \frac{1}{3u^3} + C=−(−5u−5−−3u−3)=5u51−3u31+C
Substitute back u=cosxu = \cos xu=cosx: =15cos5x−13cos3x+C= \frac{1}{5\cos^5 x} – \frac{1}{3\cos^3 x} + C=5cos5x1−3cos3x1+C
5. ∫x2−9×3 dx\int \frac{\sqrt{x^2 – 9}}{x^3}\, dx∫x3x2−9dx
Solution:
Use trigonometric substitution:
Let x=3secθ⇒dx=3secθtanθ dθx = 3\sec\theta \Rightarrow dx = 3\sec\theta\tan\theta\, d\thetax=3secθ⇒dx=3secθtanθdθ
Then, x2−9=9sec2θ−9=9tan2θ=3tanθ\sqrt{x^2 – 9} = \sqrt{9\sec^2\theta – 9} = \sqrt{9\tan^2\theta} = 3\tan\thetax2−9=9sec2θ−9=9tan2θ=3tanθ
Now substitute: ∫3tanθ(27sec3θ)⋅3secθtanθ dθ=∫9tan2θ27sec2θdθ=13∫tan2θcos2θ dθ\int \frac{3\tan\theta}{(27\sec^3\theta)} \cdot 3\sec\theta\tan\theta\, d\theta = \int \frac{9\tan^2\theta}{27\sec^2\theta} d\theta = \frac{1}{3} \int \tan^2\theta \cos^2\theta\, d\theta∫(27sec3θ)3tanθ⋅3secθtanθdθ=∫27sec2θ9tan2θdθ=31∫tan2θcos2θdθ
But tan2θ=sec2θ−1\tan^2\theta = \sec^2\theta – 1tan2θ=sec2θ−1
So: =13∫(sec2θ−1)cos2θ dθ=13∫(1−cos2θ) dθ=13∫sin2θ dθ= \frac{1}{3} \int (\sec^2\theta – 1)\cos^2\theta\, d\theta = \frac{1}{3} \int (1 – \cos^2\theta)\, d\theta = \frac{1}{3} \int \sin^2\theta\, d\theta=31∫(sec2θ−1)cos2θdθ=31∫(1−cos2θ)dθ=31∫sin2θdθ
Use identity sin2θ=1−cos(2θ)2\sin^2\theta = \frac{1 – \cos(2\theta)}{2}sin2θ=21−cos(2θ)
Then: =13⋅12∫(1−cos(2θ)) dθ=16(θ−sin(2θ)2)+C= \frac{1}{3} \cdot \frac{1}{2} \int (1 – \cos(2\theta))\, d\theta = \frac{1}{6} (\theta – \frac{\sin(2\theta)}{2}) + C=31⋅21∫(1−cos(2θ))dθ=61(θ−2sin(2θ))+C
Back-substitute θ=sec−1(x/3)\theta = \sec^{-1}(x/3)θ=sec−1(x/3), and convert to xxx.
