∫cos(x)sin2(x)dx∫cos(x)sin2(x)dx Evaluate the indefinite integral.

# ∫cos(x)sin2(x)dx∫cos(x)sin2(x)dx Evaluate the indefinite integral.

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x=2y2,x=4+y2x=2y2,x=4+y2 Sketch the region enclosed by the given curves and find its area.

x=2y2,x=4+y2x=2y2,x=4+y2 2y2=4+y22y2=4+y2 2y2−y2=42y2-y2=4 y2=4y2=4 y=±2y=±2 Refer the attached image. Graph of x=2y^2 is plotted in red color and graph of x=4+y^2 is plotted in blue color. Area of the region…

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y=cos(x),y=2−cos(x),0≤x≤2πy=cos(x),y=2-cos(x),0≤x≤2π Sketch the region enclosed by the given curves…

y=cos(x),y=2−cos(x),0≤x≤2πy=cos(x),y=2-cos(x),0≤x≤2π Refer the attached image. Graph of y=cos(x) is plotted in red color and graph of y=2-cos(x) is plotted in blue color. Area of the region enclosed by the…

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y=ex,y=x(ex),x=0y=ex,y=x(ex),x=0 Sketch the region enclosed by the given curves and find its area.

Graph each equation to determine the bounded region. ( Red curve is the graph of y=exy=ex . Blue curve is the graph of y=xexy=xex . And green line is the graph of x=0.) Base on the graph, the…

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y=x2,y=4x−x2y=x2,y=4x-x2 Sketch the region enclosed by the given curves and find its area.

You need to determine first the points of intersection between curves y=x2y=x2 and y=4x−x2y=4x-x2 , by solving the equation, such that: x2=4x−x2⇒2×2−4x=0x2=4x-x2⇒2×2-4x=0 Factoring out 2x…

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y=12−x2,y=x2−6y=12-x2,y=x2-6 Sketch the region enclosed by the given curves and find its area.

y=12−x2,y=x2−6y=12-x2,y=x2-6 Refer the attached image. Graph of y=12-x^2 is plotted in blue color and graph of y=x^2-6 is plotted in red color. The curves intersects at x=±± 3 Area of the region enclosed…

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4x+y2=12,x=y4x+y2=12,x=y Sketch the region enclosed by the given curves. Decide whether to…

You need to determine first the points of intersection between curves y=12−4x−−−−−−√y=12-4x and y=xy=x , by solving the equation, such that: 12−4x−−−−−−√=x⇒12−4x=x2⇒x2+4x−12=012-4x=x⇒12-4x=x2⇒x2+4x-12=0…

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x=1−y2,x=y2−1x=1-y2,x=y2-1 Sketch the region enclosed by the given curves. Decide whether to…

Refer the attached image, Graph of x=1-y^2 is plotted in blue color and graph of x=y^2-1 is plotted in red color. The curves intersects at y=±± 1 Area of the region enclosed by the given…

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y=sin(x),y=2xπ,x⇒0y=sin(x),y=2xπ,x⇒0 Sketch the region enclosed by the given curves. Decide…

Here is the sketch of the two given functions. The y=sin(x)y=sin(x) is plotted with a red color while y=2xπy=2xπ is plotted with a blue color. As shown the graph, the two graphs intersect at the…

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y=1x,y=1×2,x=2y=1x,y=1×2,x=2 Sketch the region enclosed by the given curves. Decide whether…

y=1x,y=1×2,x=2y=1x,y=1×2,x=2 Refer the attached image, y=1/x is plotted in red color and y=1/x^2 is plotted in blue color. Curves intersect at x=y=1 Area of the region enclosed by the given curves…

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y=x2−2x,y=x+4y=x2-2x,y=x+4 Sketch the region enclosed by the given curves. Decide whether to…

Since the curves are y=x2−2xy=x2-2x and y = x + 4, you must integrate y with respect to x. First, you need to find the point of intersection between the curves y=x2−2xy=x2-2x and y = x + 4, by…

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y=(x−2)2,y=xy=(x-2)2,y=x Sketch the region enclosed by the given curves. Decide whether to…

I’ll integrate with respect to x. The limits of integration are x=1 and x=4, and between them x>(x−2)2.x>(x-2)2. The width of a rectangle is dxdx and the height is x−(x−2)2.x-(x-2)2. The area is

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y=sin(x),y=x,x=π2,x=πy=sin(x),y=x,x=π2,x=π Sketch the region enclosed by the given curves. Decide…

y=sin(x),y=x,x=π2,x=πy=sin(x),y=x,x=π2,x=π Refer the attached image. Graph of y=sin(x) is plotted in red color and y=x is plotted in blue color. Area of the region enclosed by the given curves…

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y=ex,y=x2−1,x=−1,x=1y=ex,y=x2-1,x=-1,x=1 Sketch the region enclosed by the given curves. Decide…

y=ex,y=x2−1,x=−1,x=1y=ex,y=x2-1,x=-1,x=1 Refer the attached image. Graph of e^x is plotted in red color and y=x^2-1 is plotted in blue color. From the graph ,the region of y=e^x lies above the region of…

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∫10dx(1+x−−√)4∫01dx(1+x)4 Evaluate the definite integral.

You need to solve the definite integral, using fundamental theorem of calculus, such that: ∫baf(x)dx=F(b)−F(a)∫abf(x)dx=F(b)-F(a) First, you need to solve the indefinite integral

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∫T20sin(2πtT−α)dt∫0T2sin(2πtT-α)dt Evaluate the definite integral.

The indefinite integral is −(T2π)⋅cos(2π⋅tT−α)+C,-(T2π)⋅cos(2π⋅tT-α)+C,therefore the definite integral is −(T2π)⋅(cos(π−α)−cos(−α))=-(T2π)⋅(cos(π-α)-cos(-α))= =(Tπ)⋅cos(α).=(Tπ)⋅cos(α).

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∫10ez+1ez+zdz∫01ez+1ez+zdz Evaluate the definite integral.

Given: ∫10ez+1ez+zdz∫01ez+1ez+zdz Evaluate the integral using the Substitution Rule. Let u=ez+zu=ez+z dudz=ez+1dudz=ez+1 dz=duez+1dz=duez+1 =∫10ez+1u⋅duez+1=∫01ez+1u⋅duez+1 =∫10(1u)du=∫01(1u)du =lnu=lnu…

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∫120sin−1(x)1−x2−−−−−√dx∫012sin-1(x)1-x2dx Evaluate the definite integral.

The indefinite integral is  Therefore the definite integral is

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∫e4edxxln(x)−−−−√∫ee4dxxln(x) Evaluate the definite integral.

You need to use the following substitution ln x=u, such that: lnx=u⇒dxx=dulnx=u⇒dxx=du ∫e4edxx⋅lnx−−−√=∫u2u1duu−−√∫ee4dxx⋅lnx=∫u1u2duu

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∫40×1+2x−−−−−−√dx∫04×1+2xdx Evaluate the definite integral.

Make the substitution 1+2x=u,1+2x=u, then x=u−12x=u-12 and dx=du2.dx=du2.The indefinite integral is ∫12⋅u−12u−−√du=(14)⋅∫(u−−√−1u−−√)du=(14)⋅((23)u32−2u12)+C=∫12⋅u-12udu=(14)⋅∫(u-1u)du=(14)⋅((23)u32-2u12)+C=…

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∫21xx−1−−−−−√dx∫12xx-1dx Evaluate the definite integral.

You need to use the following substitution x−1=ux-1=u , such that: x−1=u⇒(dx)=dux-1=u⇒(dx)=du ∫21x⋅x−1−−−−−√dx=∫u2u1(u+1)⋅u12du∫12x⋅x-1dx=∫u1u2(u+1)⋅u12du

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∫π3−π3x4sin(x)dx∫-π3π3x4sin(x)dx Evaluate the definite integral.

You need to solve the definite integral, using fundamental theorem of calculus, such that: ∫baf(x)dx=F(b)−F(a)∫abf(x)dx=F(b)-F(a) First, you need to solve the indefinite integral ∫x4⋅sinxdx∫x4⋅sinxdx ,…

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∫a0xx2+a2−−−−−−√dx∫0axx2+a2dx Evaluate the definite integral.

The indefinite integral is (12)⋅∫(x2+a2−−−−−−√)d(x2+a2)=(12)⋅∫(x2+a2)d(x2+a2)= =(12)⋅(23)⋅(x2+a2)32+C.=(12)⋅(23)⋅(x2+a2)32+C. Therefore the definite integral is (13)⋅((2a2)32−(a2)32)=(13)⋅a3⋅(232−1).(13)⋅((2a2)32-(a2)32)=(13)⋅a3⋅(232-1).

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∫a0xa2−x2−−−−−−√dx∫0axa2-x2dx Evaluate the definite integral.

The indefinite integral is −(12)⋅∫(a2−x2−−−−−−√)d(a2−x2)=-(12)⋅∫(a2-x2)d(a2-x2)= =−(12)⋅(23)⋅(a2−x2)32+C.=-(12)⋅(23)⋅(a2-x2)32+C. Therefore the definite integral is −(13)⋅(0−(a2)32)=a33.-(13)⋅(0-(a2)32)=a33.

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∫130dx1+2x−−−−−−√32∫013dx1+2×32 Evaluate the definite integral.

The indefinite integral is ∫(12)⋅d(1+2x)(1+2x)23=(12)⋅∫(1+2x)−23d(1+2x)=∫(12)⋅d(1+2x)(1+2x)23=(12)⋅∫(1+2x)-23d(1+2x)= =(12)⋅3⋅(1+2x)13+C.=(12)⋅3⋅(1+2x)13+C. Therefore the definite integral is (32)⋅(2713−1)=(32)⋅(3−1)(32)⋅(2713-1)=(32)⋅(3-1) =3.

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∫π20cos(x)sin(sin(x))dx∫0π2cos(x)sin(sin(x))dx Evaluate the definite integral.

The indefinite integral is ∫(sin(sin(x))d(sin(x))=−cos(sin(x))+C.∫(sin(sin(x))d(sin(x))=-cos(sin(x))+C. Therefore the definite integral is −cos(1)+cos(0)=1−cos(1)≈0.46.-cos(1)+cos(0)=1-cos(1)≈0.46.

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∫π4−π4(x3+x4tan(x))dx∫-π4π4(x3+x4tan(x))dx Evaluate the definite integral.

You may start by checking if the function f(x)=x3+x4⋅tanxf(x)=x3+x4⋅tanx is an odd or even function, such that:

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∫10xe−x2dx∫01xe-x2dx Evaluate the definite integral.

The indefinite integral is (12)⋅∫e−x2d(x2)=−(12)⋅e−x2+C.(12)⋅∫e-x2d(x2)=-(12)⋅e-x2+C. Therefore the definite integral is −(12)⋅(e−1−1)=1−1e2≈0.316.-(12)⋅(e-1-1)=1-1e2≈0.316.

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∫21(e1xx2)dx∫12(e1xx2)dx Evaluate the definite integral.

Given: ∫21e1xx2dx∫12e1xx2dx Integrate by using u-substitution. Let u=1xu=1x u=x−1u=x-1 dudx=−1x2dudx=-1×2 dx=−x2dudx=-x2du =∫21eux2⋅(−x2)du=∫12eux2⋅(-x2)du =−∫21eudu=-∫12eudu =−eu=-eu Evaluated from x=1 to…

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∫1216csc(πt)cot(πt)dt∫1612csc(πt)cot(πt)dt Evaluate the definite integral.

Given ∫1216csc(πt)cot(πt)dt∫1612csc(πt)cot(πt)dt Integrate using the Substitution Rule. Let u=πtu=πt dudt=πdudt=π dt=duπdt=duπ =∫1216csc(u)cot(u)⋅duπ=∫1612csc(u)cot(u)⋅duπ…

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∫π0sec2(t4)dt∫0πsec2(t4)dt Evaluate the definite integral.

Given ∫π0sec2(t4)dt∫0πsec2(t4)dt Integrate using the Substitution Rule. Let u=t4u=t4 dudt=14dudt=14 dt=4dudt=4du =∫π0sec2(u)⋅4du=∫0πsec2(u)⋅4du =4∫π0sec2(u)du=4∫0πsec2(u)du =4⋅tan(u)=4⋅tan(u) evaluated from t=0 to t=ππ…

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∫30dx5x+1∫03dx5x+1 Evaluate the definite integral.

The indefinite integral is (1/5)*ln|5x+1|+C, therefore the definite integral is equal to (1/5)*(ln(16)-ln(1))=ln(16)/5 approx 0.55.

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∫101+7x−−−−−−√3dx∫011+7x3dx Evaluate the definite integral.

Given ∫101+7x−−−−−−√3dx∫011+7x3dx Integrate using the u-substitution method. Let u=1+7xu=1+7x dudx=7dudx=7 dx=du7dx=du7 =∫10u13⋅du7=∫01u13⋅du7 =17∫10u13du=17∫01u13du =17⋅u4343=17⋅u4343 Evaluated from…

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∫10(3t−1)50dt∫01(3t-1)50dt Evaluate the definite integral.

Given: ∫10(3t−1)50dt∫01(3t-1)50dt Integrate using the Substitution Rule. Let u=3t−1u=3t-1 dudt=3dudt=3 dt=du3dt=du3 =∫10u50⋅du3=∫01u50⋅du3 =13∫10u50du=13∫01u50du =13⋅u5151=13⋅u5151 evaluated from t=0 to t=1…

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∫10cos(πt2)dt∫01cos(πt2)dt Evaluate the definite integral.

Given: ∫10cos(πt2)dt∫01cos(πt2)dt Integrate using the u-substitution method. Let u=π2tu=π2t dudt=π2dudt=π2 dt=2π⋅dudt=2π⋅du =∫10cos(u)⋅2πdu=∫01cos(u)⋅2πdu =2π∫10cos(u)du=2π∫01cos(u)du =2πsin(u)=2πsin(u) Evaluated from…

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∫x3x2+1−−−−−√dx∫x3x2+1dx Evaluate the indefinite integral.

You need to use the following substitution x2+1=ux2+1=u , such that: x2+1=u⇒2xdx=du⇒xdx=du2x2+1=u⇒2xdx=du⇒xdx=du2 x2=u−1×2=u-1 ∫x3⋅x2+1−−−−−√dx=(12)∫(u−1)⋅u−−√⋅du∫x3⋅x2+1dx=(12)∫(u-1)⋅u⋅du

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∫x(2x+5)8dx∫x(2x+5)8dx Evaluate the indefinite integral.

Since it would be very hard to raise to the 8th power the binomial 2x + 5, you need to use the following substitution 2x+5=u2x+5=u , such that: 2x+5=u⇒2dx=du⇒dx=du22x+5=u⇒2dx=du⇒dx=du2 x=u−52x=u-52…

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∫x22+x−−−−−√dx∫x22+xdx Evaluate the indefinite integral.

You need to evaluate the indefinite integral ∫x2⋅2+x−−−−−√dx∫x2⋅2+xdx using the following substitution 2+x=u2+x=u , such that: 2+x=u⇒dx=du2+x=u⇒dx=du x=u−2x=u-2

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∫1+x1+x2dx∫1+x1+x2dx Evaluate the indefinite integral.

You need to evaluate the indefinite integral ∫1+x1+x2dx∫1+x1+x2dx such that: ∫1+x1+x2dx=∫11+x2dx+∫x1+x2dx∫1+x1+x2dx=∫11+x2dx+∫x1+x2dx ∫1+x1+x2dx=arctanx+∫x1+x2dx∫1+x1+x2dx=arctanx+∫x1+x2dx You need…

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∫x1+x4dx∫x1+x4dx Evaluate the indefinite integral.

You need to use the following substitution x2=ux2=u , such that: x2=u⇒2xdx=du⇒xdx=du2x2=u⇒2xdx=du⇒xdx=du2 ∫x⋅dx1+x4=(12)⋅∫du1+u2∫x⋅dx1+x4=(12)⋅∫du1+u2 (12)⋅∫du1+u2=(12)⋅arctanu+c(12)⋅∫du1+u2=(12)⋅arctanu+c…

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∫dx1−x2−−−−−√sin−1(x)∫dx1-x2sin-1(x) Evaluate the indefinite integral.

You need to evaluate the indefinite integral, using the following substitution, such that: sin−1x=t⇒dx1−x2−−−−−√=dtsin-1x=t⇒dx1-x2=dt Replacing t for x yields:

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∫sin(t)sec2(cos(t))dt∫sin(t)sec2(cos(t))dt Evaluate the indefinite integral.

You need to use the following substitution cost=u,cost=u, such that: cost=u⇒−sintdt=du⇒sintdt=−ducost=u⇒-sintdt=du⇒sintdt=-du ∫sint⋅sec2(cost)dt=−∫sec2udu∫sint⋅sec2(cost)dt=-∫sec2udu −∫sec2udu=−tanu+c-∫sec2udu=-tanu+c…

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∫cot(x)dx∫cot(x)dx Evaluate the indefinite integral.

Given ∫cot(x)dx∫cot(x)dx =∫cos(x)sin(x)dx=∫cos(x)sin(x)dx  Let u=sin(x)u=sin(x) dudx=cos(x)dudx=cos(x) dx=ducos(x)dx=ducos(x) =∫cos(x)uducos(x)=∫cos(x)uducos(x) =∫1udu=∫1udu =ln|u|+C=ln|u|+C =ln|sin(x)|+C=ln|sin(x)|+C

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∫sin(x)1+cos2(x)dx∫sin(x)1+cos2(x)dx Evaluate the indefinite integral.

Given ∫sin(x)1+cos2(x)dx∫sin(x)1+cos2(x)dx Integrate using the Substitution Rule. Let u=cos(x)u=cos(x) dudx=−sin(x)dudx=-sin(x) dx=du−sin(x)dx=du-sin(x) =∫sin(x)1+u2dx=∫sin(x)1+u2dx =∫sin(x)1+u2⋅du−sin(x)=∫sin(x)1+u2⋅du-sin(x)…

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∫sin(2x)1+cos2(x)dx∫sin(2x)1+cos2(x)dx Evaluate the indefinite integral.

Given ∫sin(2x)1+cos2(x)dx∫sin(2x)1+cos2(x)dx From trigonometry sin(2x)=2sin(x)cos(x)sin(2x)=2sin(x)cos(x) =∫2sin(x)cos(x)1+cos2(x)dx=∫2sin(x)cos(x)1+cos2(x)dx =2∫sin(x)cos(x)1+cos2(x)dx=2∫sin(x)cos(x)1+cos2(x)dx Integrate using u-substitution. Let…

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∫dtcos2(t)1+tan(t)−−−−−−−−√∫dtcos2(t)1+tan(t) Evaluate the indefinite integral.

∫dtcos2(t)1+tan(t)−−−−−−−−√∫dtcos2(t)1+tan(t) =∫sec2(t)1+tan(t)−−−−−−−−√dt=∫sec2(t)1+tan(t)dt apply integral substitution, let x=1+tan(t) dx=sec2(t)dtdx=sec2(t)dt =∫dxx−−√=∫dxx =x−12+1−12+1=x-12+1-12+1 =x1212=x1212…

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∫sinh2(x)cosh(x)dx∫sinh2(x)cosh(x)dx Evaluate the indefinite integral.

You need to evaluate the indefinite integral by using the substitution sinhx=tsinhx=t , such that: sinhx=t⇒coshxdx=dtsinhx=t⇒coshxdx=dt ∫sinh2x⋅coshxdx=∫t2dt∫sinh2x⋅coshxdx=∫t2dt Using the formula

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∫2t2t+3dt∫2t2t+3dt Evaluate the indefinite integral.

You need to use the following substitution 2t+3=u2t+3=u , such that: 2t+3=u⇒2t⋅ln2dt=du⇒2t⋅dt=duln22t+3=u⇒2t⋅ln2dt=du⇒2t⋅dt=duln2 ∫2t⋅dt2t+3=(1ln2)⋅∫duu∫2t⋅dt2t+3=(1ln2)⋅∫duu

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∫cot(x)−−−−−√csc2(x)dx∫cot(x)csc2(x)dx Evaluate the indefinite integral.

You need to use the following substitution cotx=tcotx=t , such that: cotx=t⇒−csc2xdx=dt⇒csc2xdx=−dtcotx=t⇒-csc2xdx=dt⇒csc2xdx=-dt ∫cotx−−−−√⋅csc2xdx=−∫t√dt∫cotx⋅csc2xdx=-∫tdt

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∫cos(πx)x2dx∫cos(πx)x2dx Evaluate the indefinite integral.

Given ∫cos(πx)x2dx∫cos(πx)x2dx integrate using the Substitution Rule. Let u=πxu=πxor u=πx−1u=πx-1 dudx=−πx−2dudx=-πx-2 dudx=−πx2dudx=-πx2 dx=x2−π⋅dudx=x2-π⋅du =∫cos(u)x2⋅(x2−π)du=∫cos(u)x2⋅(x2-π)du…