Evaluate the following integrals:
Question: Evaluate $\int \frac{x^{2}}{\left(1+x^{2}\right)^{2}} d x$ Solution: Let, $x=\tan t$ Differentiating both side with respect to t $\frac{d x}{d t}=\sec ^{2} t \Rightarrow d x=\sec ^{2} t d t$ $y=\int \frac{\tan ^{3} t}{\sec ^{4} t} \sec ^{2} t d t$ $y=\int \frac{\sin ^{3} t}{\cos t} d t$ $y=\int \frac{\left(1-\cos ^{2} t\right) \sin t}{\cos t} d t$ Again, let $\cos t=z$ Differentiating both side with respect to t $\frac{d z}{d t}=-\sin t \Rightarrow-d z=\sin t d t$ $y=-\int \frac{\left(...
Read More →Prove the following equation
Question: $\int \tan ^{2} x \sec ^{4} x d x$ Solution: Let $\mathrm{I}=\int \tan ^{2} x \cdot \sec ^{4} x d x$ $=\int \tan ^{2} x \sec ^{2} x \cdot \sec ^{2} x d x=\int \tan ^{2} x\left(1+\tan ^{2} x\right) \cdot \sec ^{2} x d x$ Putting, $\tan x=t, \Rightarrow \sec ^{2} x d x=d t$ $I=\int t^{2}\left(1+t^{2}\right) d t=\int\left(t^{2}+t^{4}\right) d t=\int t^{2} d t+\int t^{4} d t$ $=\frac{1}{3} t^{3}+\frac{1}{5} t^{5}=\frac{1}{3} \tan ^{3} x+\frac{1}{5} \tan ^{5} x+C$ Therefore, $\mathrm{I}=\fr...
Read More →Evaluate the following integrals:
Question: Evaluate $\int x \sqrt{2 x+3} d x$ Solution: In this question we write $x \sqrt{2 x+3}$ as $x \sqrt{2 x+3}=\frac{2 x \sqrt{2 x+3}}{2}$ $=\frac{(2 x+3-3) \sqrt{2 x+3}}{2}$ $=\frac{(2 x+3) \sqrt{2 x+3}-3 \sqrt{2 x+3}}{2}$ $=\frac{(2 x+3)^{\frac{3}{2}}-3 \sqrt{2 x+3}}{2}$ $y=\int x \sqrt{2 x+3} d x$ $y=\int \frac{(2 x+3)^{\frac{3}{2}}-3 \sqrt{2 x+3}}{2} d x$ Using formula $\int(a x+b)^{n} d x=\frac{(a x+b)^{n+1}}{a(n+1)}$ $y=\frac{(2 x+3)^{\frac{5}{2}}}{2 \times 2 \times \frac{5}{2}}-\fra...
Read More →The value of
Question: $\int \frac{d x}{1+\cos x}$ Solution: Let $\mathrm{I}=\int \frac{d x}{1+\cos x}=\int \frac{d x}{2 \cos ^{2} x / 2} \quad\left[\because 1+\cos x=2 \cos ^{2} \frac{x}{2}\right]$ $=\frac{1}{2} \int \sec ^{2} \frac{x}{2} d x=\frac{1}{2} \cdot 2 \tan \frac{x}{2}+C=\tan \frac{x}{2}+C$ Therefore, $I=\tan \frac{x}{2}+C$...
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Question: $\int \frac{(1+\cos x)}{x+\sin x} d x$ Solution: $I=\int \frac{1+\cos x}{x+\sin x} d x$ Putting, $x+\sin x=t \Rightarrow(1+\cos x) d x=d t$ So, $I=\int \frac{d t}{t}=\log |t|=\log |x+\sin x|+C$ Therefore, $\mathrm{I}=\log |t|=\log |x+\sin x|+\mathrm{C}$...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{x^{2}}{(x-1)^{2}} d x$ Solution: $y=\int \frac{(x-1+1)^{2}}{(x-1)^{2}} d x$ $y=\int \frac{(x-1)^{2}+2(x-1)+1}{(x-1)^{3}} d x$ $y=\int \frac{1}{(x-1)}+2 \frac{1}{(x-1)^{2}}+\frac{1}{(x-1)^{3}} d x$ Using formula $\int \frac{1}{x} d x=\ln x$ and $\int x^{n} d x=\frac{x^{n+1}}{n+1}$ $y=\ln (x-1)+2 \frac{(x-1)^{-1}}{-1}+\frac{(x-1)^{-2}}{-2}+c$ $y=\ln (x-1)-2(x-1)^{-1}-\frac{(x-1)^{-2}}{2}+c$...
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Question: $\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} d x$ Solution: Let $\mathrm{I}=\int \frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}} d x=\int \frac{e^{\log x^{4}}-e^{\log x^{3}}}{e^{\log x^{6}}-e^{\log x^{9}}} d x$ $=\int \frac{x^{6}-x^{5}}{x^{4}-x^{3}} d x$ (Using properties of logarithms) $=\int \frac{x^{2}\left(x^{4}-x^{3}\right)}{x^{4}-x^{3}} d x=\int x^{2} d x=\frac{1}{3} x^{3}+C$ Therefore, $\mathrm{I}=\int x^{2} d x=\frac{1}{3} x^{3}+\mathrm{C}$...
Read More →Evaluate the following:
Question: Evaluate the following: $\int \frac{\left(x^{2}+2\right) d x}{x+1}$ Solution: $\mathrm{I}=\frac{x^{2}}{2}-x+3 \log |x+1|+\mathrm{C}$...
Read More →Prove the following
Question: $\int \frac{2 x+3}{x^{2}+3 x} d x=\log \left|x^{2}+3 x\right|+C$ Solution: L.H.S. $=\int \frac{2 x+3}{x^{2}+3 x} d x$ Putting, $\quad x^{2}+3 x=t$ So, $\quad(2 x+3) d x=d t$ $\Rightarrow \quad \int \frac{d t}{t}=\log |t| \Rightarrow \log \left|x^{2}+3 x\right|+\mathrm{C}=$ R.H.S. L.H.S. $=$ R.H.S - Hence proved....
Read More →Verify the following
Question: Verify the following $\int \frac{2 x-1}{2 x+3} d x=x-\log \left|(2 x+3)^{2}\right|+C$ Solution: L.H.S. $=\int \frac{2 x-1}{2 x+3} d x$ $=\int\left(1-\frac{4}{2 x+3}\right) d x$ [Dividing the numerator by the denominator] $=\int 1 \cdot d x-4 \int \frac{1}{2 x+3} d x=\int 1 \cdot d x-\frac{4}{2} \int \frac{1}{x+\frac{3}{2}} d x$ $=\int 1 \cdot d x-2 \int \frac{1}{x+\frac{3}{2}} d x=x-2 \log \left|x+\frac{3}{2}\right|+C$ $=x-2 \log \left|\frac{2 x+3}{2}\right|+C=x-\log \left|\left(\frac{...
Read More →AB is a diameter of a circle and C is any point on the circle.
Question: AB is a diameter of a circle and C is any point on the circle. Show that the area of Δ ABC is maximum, when it is isosceles. Solution: Let consider AB be the diameter and C is any point on the circle with radius r. ACB = 90o[angle in the semi-circle is 90o] Let AC = x $B C=\sqrt{\bar{A} B^{2}-A C^{2}}$ $\mathrm{BC}=\sqrt{(2 r)^{2}-x^{2}} \Rightarrow \mathrm{BC}=\sqrt{4 r^{2}-x^{2}}$$\ldots(i)$ Now area of $\triangle \mathrm{ABC}, \mathrm{A}=\frac{1}{2} \times \mathrm{AC} \times \mathrm...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{6 x+5}{\sqrt{6+x-2 x^{2}}} d x$ Solution: $y=6 \int \frac{x+\frac{5}{6}}{\sqrt{6+x-2 x^{2}}} d x$ $y=\frac{6}{-4} \int \frac{-4\left(x+\frac{5}{6}\right)}{\sqrt{6+x-2 x^{2}}} d x$ $y=-\frac{3}{2} \int \frac{-4 x-\frac{10}{3}+1-1}{\sqrt{6+x-2 x^{2}}} d x$ $y=-\frac{3}{2} \int \frac{-4 x+1}{\sqrt{6+x-2 x^{2}}} d x-\frac{3}{2} \int \frac{-\frac{10}{3}-1}{\sqrt{6+x-2 x^{2}}} d x$ $A=-\frac{3}{2} \int \frac{-4 x+1}{\sqrt{6+x-2 x^{2}}} d x$ Let, $6+x-2 x^{2}=t$ Different...
Read More →If the sum of the surface areas of cube and a sphere is constant,
Question: If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum? Solution: sphere. Surface area of cube = 6x2 And, surface area of the sphere = 4r2 Now, their sum is $6 x^{2}+4 \pi r^{2}=\mathrm{K}($ constant $) \Rightarrow r=\sqrt{\frac{\mathrm{K}-6 x^{2}}{4 \pi}}$.............(i) Volume of the cube $=x^{3}$ and the volume of sphere $=\frac{4}{3} \pi r^{3}$ Now, Sum of th...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \sqrt{\frac{a+x}{x}} d x$ Solution: to solve this integral we have to apply substitution method for which we are going to put $x=a \cdot \tan ^{2} k$ This means $\mathrm{dx}=2$.a.tank.sec ${ }^{2} k \cdot d k$, then I will be, $I=\int \sqrt{\frac{\operatorname{asec}^{2} k}{a \tan ^{2} k}} \cdot 2 a \cdot \tan k \cdot \sec ^{2} k \cdot d k=2 a \cdot \operatorname{cosec} k \cdot \tan k \cdot \sec ^{2} k \cdot d k$ In this above integral let $\tan k=t$ then $\sec ^{2} k d k...
Read More →Find the dimensions of the rectangle of perimeter 36 cm
Question: Find the dimensions of the rectangle of perimeter 36 cm which will sweep out a volume as large as possible, when revolved about one of its sides. Also find the maximum volume. Solution: Lets consider x and y to be the length and breadth of given rectangle ABCD. According to the question, the rectangle will be resolved about side AD which making a cylinder with radius x and height y. So, the volume of the cylinder V = r2h = x2y . (1) 36 = 2(x + y) 18 = x + y y = 18 x .. (ii) Putting the...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{1}{2+\cos x} d x$ Solution: To solve this type of solution , we are going to substitute the value of $\sin x$ and $\cos x$ in terms of $\tan (x / 2)$ $\sin x=\frac{2\left[\tan \frac{x}{2}\right]}{1+\tan ^{2} \frac{x}{2}}$ $\cos x=\frac{\left(1-\frac{\tan ^{2} x}{2}\right)}{1+\frac{\tan ^{2} x}{2}}$ $I=\int \frac{1}{\left(2+\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$ $I=\int \frac{\sec ^{2} \frac{x}{2}}{\left(2+2 \tan ^{2} \frac{x}{2}+1-\tan...
Read More →An open box with square base is to be made of a given quantity
Question: An open box with square base is to be made of a given quantity of card board of areac2. Show that the maximum volume of the box is c3/ 63 cubic units. Solution: Let x be the length of the side of the square base of the cubical open box and y be its height. So, the surface area of the open box $c^{2}=x^{2}+4 x y \Rightarrow y=\frac{c^{2}-x^{2}}{4 x}$ $\ldots(i)$ Now volume of the box, $\mathrm{V}=x \times x \times y$ $V=x^{2} y$ $\mathrm{V}=x^{2}\left(\frac{c^{2}-x^{2}}{4 x}\right)$ $\R...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{1+\sin x}{\sin x(1+\cos x)} d x$ Solution: first divide nominator by denominator - $I=\int \frac{1}{\sin x(1+\cos x)} d x+\int \frac{1}{1+\cos x} d x$ $=\int \frac{1}{\sin x(1+\cos x)} d x+\int \frac{1}{1+2 \cos ^{2} x-1} d x$ : To solve this type of solution, we are going to substitute the value of $\sin x$ and $\cos x$ in terms of tan( $x / 2)$ $\sin x=\frac{2\left[\tan \frac{x}{2}\right]}{1+\tan ^{2} \frac{x}{2}}$ $\cos x=\frac{\left(1-\frac{\tan ^{2} x}{2}\righ...
Read More →If the straight-line x cos α + y sin α = p touches the curve
Question: If the straight-linexcos +ysin =ptouches the curve x2/a2+ y2/b2= 1, then prove thata2cos2 +b2sin2 =p2. Solution: The given curve is x2/a2+ y2/b2= 1 and the straight-linexcos +ysin =p Differentiating equation (i) w.r.t. x, we get $\frac{1}{a^{2}} \cdot 2 x+\frac{1}{b^{2}} \cdot 2 y \cdot \frac{d y}{d x}=0$ $\Rightarrow \frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{b^{2}}{a^{2}} \cdot \frac{x}{y}$ So the slope of the curve $=\frac{-b^{2}}{a^{2}} \cd...
Read More →A telephone company in a town has 500 subscribers
Question: A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs 300/- per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of Rs 1/- one subscriber will discontinue the service. Find what increase will bring maximum profit? Solution: Lets consider that the company increases the annual subscription by Rs x. So, x is the number of subscribers who discontinue the services. Total revenue,...
Read More →Find the points of local maxima,
Question: Find the points of local maxima, local minima and the points of inflection of the functionf(x) =x5 5x4+ 5x3 1. Also find the corresponding local maximum and local minimum values. Solution: Given,f(x) =x5 5x4+ 5x3 1 Differentiating the function, f(x) = 5x4 20x3+ 15x2 For local maxima and local minima,f(x) = 0 $5 x^{4}-20 x^{3}+15 x^{2}=0 \Rightarrow 5 x^{2}\left(x^{2}-4 x+3\right)=0$ $\Rightarrow 5 x^{-2}\left(x^{2}-3 x-x+3\right)=0 \Rightarrow x^{2}(x-3)(x-1)=0$ $\therefore x=0, x=1$ a...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \operatorname{cosec}^{4} 2 x d x$ Solution: $I=\int \operatorname{cosec}^{4} 2 x d x$ $=\int \operatorname{cosec}^{2} 2 x\left(1+\cot ^{2} 2 x\right) d x$ $=\int \operatorname{cosec}^{2} 2 x d x+\int \operatorname{cosec}^{2} 2 x \cot ^{2} 2 x d x$ Let us consider that $\cot 2 x=t$ then $-2 \cdot \operatorname{cosec}^{2} 2 x d x=d t$ $I=-\frac{\cot (2 x)}{2}-\frac{1}{2} \cdot\left(t^{2} d t\right)$ $I=-\frac{\cot (2 x)}{2}-\frac{1}{6} \cdot(\cot 2 x)^{3}$...
Read More →If the sum of the lengths of the hypotenuse and a side of a right angled
Question: If the sum of the lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum when the angle between them is /3. Solution: Let AC = x, BC = y So, AB = (x2+ y2) ACB = Let z = x + y (given) Now, the area of ∆ABC = x AB x BC $\Rightarrow \mathrm{A}=\frac{1}{2} y \cdot \sqrt{x^{2}-y^{2}} \Rightarrow \mathrm{A}=\frac{1}{2} y \cdot \sqrt{(Z-y)^{2}-y^{2}}$ Squaring both sides, we get $\mathrm{A}^{2}=\frac{1}{4} y^{2}\left[(Z-y)^{2}-y...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \sec ^{4} x d x$ Solution: above equation can be solve by using one formula that is $\left(i+\tan ^{2} x=\sec ^{2} x\right)$ $I=\int \sec ^{4} x d x$ $=\int \sec ^{2} x \sec ^{2} x d x$ $=\int \sec ^{2} x\left(1+\tan ^{2} x\right) d x$ $=\int \sec ^{2} x d x+\int \sec ^{2} x \tan ^{2} x d x$ Put $\tan x=t$ in above equation so that $\sec ^{2} x d x=d t$ $I=\tan x+\int t^{2} d t=\tan x+\frac{t^{3}}{3}$ $=\tan x+\frac{\tan ^{3} x}{3}$...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{1}{5-4 \sin x} d x$ Solution: in this integral we are going to put the value of $\sin (x)$ in terms of $\tan (x / 2)$ - $\begin{aligned} I =\int \frac{2 d t}{5+5 t^{2}-8 t} \\ I =\frac{2}{5} \int \frac{1}{\left(t-\frac{4}{5}\right)^{2}+\left(\frac{3}{5}\right)^{2}} d t \end{aligned}$ By applying the formula of $1 /\left(x^{2}+a^{2}\right)$ in above equation yields the integral- $I=\frac{2}{5} \cdot \frac{1}{\frac{3}{5}} \cdot \tan ^{-1} \frac{\left(t-\frac{4}{5}\ri...
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