Question 2015 Marks
Evaluate the following integrals:
$\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big\}\text{dx}$
AnswerLet $\text{I}=\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big\}\text{dx}$
Let $\text{x}=\cos2\theta$
On differentiating both sides, we get
$\text{dx}=-2\sin2\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\cos2\theta}{1-\cos2\theta}}\Big\}2\sin2\theta\text{ d}\theta$
$=-2\int\cos\Big\{2\cot^{-1}\sqrt{\frac{2\cos^2\theta}{2\sin^2\theta}}\Big\}\sin2\theta\text{ d}\theta$
$=-2\int\cos\{2\cot^{-1}(\cot\theta)\}\sin2\theta\text{ d}\theta$
$=-2\int\cos2\theta\sin2\theta\text{ d}\theta$
$=\frac{\cos4\theta}{4}+\text{C}_1$
$=-\int\sin4\theta\text{ d}\theta$
$=\frac{1}{4}(2\cos^2\theta-1)+\text{C}_1$
$=\frac{1}{2}\text{x}^2-\frac{1}{4}+\text{C}_1$
$=\frac{1}{2}\text{x}^2+\text{C},$ where $\text{C}=-\frac{1}{4}+\text{C}_1$
Hence, $\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big\}\text{dx}=\frac{1}{2}\text{x}^2+\text{C}$
View full question & answer→Question 2025 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}$
Answer$\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}$
Let $\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}=\frac{\text{A}}{\text{x}-1}+\frac{\text{B}}{\text{x}-2}+\frac{\text{C}}{\text{x}-3}$
$\Rightarrow\frac{\text{x}^2}{(\text{x}^2-1)(\text{x}-2)(\text{x}-3)}=\frac{\text{A}(\text{x}-2)(\text{x}-3)+\text{B}(\text{x}-1)(\text{x}-3)+\text{C}(\text{x}-1)(\text{x}-2)}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}$
$\Rightarrow\text{x}^2=\text{A}(\text{x}-2) (\text{x}-3)+\text{B}(\text{x}-1)(\text{x}-3)\\+\text{C}(\text{x}-1)(\text{x}-2)\ \dots(1)$
Putting x - 1 = 0 or x = 1 in eq (1)
⇒ 1 = A (1 - 2) (1 - 3)
⇒ 1 = A (-1) (-2)
$\text{A}=\frac{1}{2}$
Putting x - 2 = 0 or x = 2 in eq (1)
⇒ 4 = B (2 - 1)(2 - 3)
⇒ B = -4
Putting x - 3 = 0 or x = 3 in eq (1)
⇒ 9 = C (3 - 1) (3 - 2)
$\Rightarrow\text{C}=\frac{9}{2}$
$\therefore\frac{\text{x}^2}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}=\frac{1}{2(\text{x}-1)}-\frac{4}{\text{x}-2}+\frac{9}{2(\text{x}-3)}$
$\int\frac{\text{x}^2}{(\text{x}-1)(\text{x}-2)(\text{x}-3)}\ \text{dx}=\frac{1}{2}\int\frac{1}{\text{x}-1}\ \text{dx}-4\int\frac{1}{\text{x}-2}+\frac{9}{2(\text{x}-3)}\ \text{dx}$
$=\frac{1}{2}\ln|\text{x}-1|-4\ln|\text{x}-2|+\frac{9}{2}\ln|\text{x}-3|+\text{C}$
View full question & answer→Question 2035 Marks
Evaluate the following integrals:
$\int\sqrt{1+\text{x}-2\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{1+\text{x}-2\text{x}^2}\text{dx}$
$=\sqrt{2}\int\sqrt{\frac{1}{2}+\frac{\text{x}}{2}-\text{x}^2}\text{dx}$
$=\sqrt{2}\int\sqrt{\frac{9}{16}-\Big(\frac{1}{16}-\frac{\text{x}}{2}+\text{x}^2\Big)}\text{dx}$
$=\sqrt{2}\int\sqrt{\Big(\frac{3}{4}\Big)^2-\Big(\text{x}-\frac{1}{4}\Big)^2}\text{dx}$
$=\sqrt{2}\begin{Bmatrix}\frac{\Big(\text{x}-\frac{1}{4}\Big)}{2}\sqrt{\frac{1}{2}+\frac{\text{x}}{2}-\text{x}^2}+\frac{9}{32}\sin^{-1}\bigg(\frac{\text{x}-\frac{1}{4}}{\frac{3}{4}}\bigg)\end{Bmatrix}+\text{C}$
$\text{I}=\frac{1}{8}(4\text{x}-1)\sqrt{1+\text{x}-2\text{x}^2}+\frac{9\sqrt{2}}{32}\sin^{-1}\Big(\frac{4\text{x}-1}{3}\Big)+\text{C}$
View full question & answer→Question 2045 Marks
Evaluate the following intregals:
$\int\frac{5\text{x}^2-1}{\text{x}(\text{x}-1)(\text{x}+1)}\ \text{dx}$
AnswerLet $\int\frac{5\text{x}^2-1}{\text{x}(\text{x}-1)(\text{x}+1)}\ \text{dx}=\frac{\text{A}}{\text{x}}+\frac{\text{B}}{\text{x}-1}+\frac{\text{C}}{\text{x}+1}$
$\Rightarrow5\text{x}^2-1=\text{A}(\text{x}^2-1)+\text{B}(\text{x}+1)\text{x}+\text{C}(\text{x}-1)\text{x}$
Put x = 0
⇒ -1 = -A ⇒ A = 1
Put x = +1
⇒ 4 = 2B ⇒ B = 2
Put x = -1
⇒ 4 = 2C ⇒ C = 2
So,
$\text{I}=\int\frac{\text{dx}}{\text{x}}+\int\frac{2\text{dx}}{\text{x}-1}+\int\frac{2\text{dx}}{\text{x}+1}$
$=\log|\text{x}|+2\log|\text{x}-1|+2\log|\text{x}+1|+\text{C}$
$\text{I}=\log|\text{x}(\text{x}^2-1)^2|$
View full question & answer→Question 2055 Marks
Evaluate the following integrals:
$\int\tan^5\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\tan^5\text{x}\text{ dx}$ Then
$\text{I}=\int\tan^2\text{x}\tan^3\text{x}\text{ dx}$
$=\int(\sec^2\text{x}-1)\tan^3\text{x}\text{ dx}$
$=\int\sec^2\text{x}\tan^3\text{x}\text{ dx}-\int\tan^3\text{x}\text{ dx}$
$=\int\sec^2\tan^3\text{x}\text{ dx}-\int(\sec^2\text{x}-1)\tan\text{x}\text{ dx}$
Substituting $\tan\text{x}=\text{t}$ and $\sec^2\text{x}\text{ dx}=\text{dt}$ in first two integral, we get
$\text{I}=\int\text{t}^3\text{dt}-\int\text{tdt}+\int\tan\text{x}\text{ dx}$
$=\frac{\text{t}^4}{4}-\frac{\text{t}^2}{2}+\log|\sec\text{x}|+\text{C}$
$=\frac{\tan^4\text{x}}{4}-\frac{\tan^2\text{x}}{2}+\log|\sec\text{x}|+\text{C}$
$\therefore\ \text{I}=\frac{\tan^4\text{x}}{4}-\frac{\tan^2\text{x}}{2}+\log|\sec\text{x}|+\text{C}$
View full question & answer→Question 2065 Marks
Evalute the following integrals:
$\int\frac{1}{\text{e}^\text{x}+1}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{e}^\text{x}+1}\text{dx}$ then,
$\text{I}=\int\frac{1}{\text{e}^\text{x}\Big[1+\frac{1}{\text{e}^\text{x}}\Big]}\text{dx}$
$\Rightarrow\text{I}=\int\frac{1}{\text{e}^\text{x}\big[1+\text{e}^{-\text{x}}\big]}\text{dx}\ .....\text{(i)}$
Let $1+\text{e}^{-\text{x}}=\text{t}$ then,
$\text{d}(1+\text{e}^{-\text{x}})=\text{dt}$
$\Rightarrow-\text{e}^{-\text{x}}\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{-\text{dt}}{\text{e}^{-\text{x}}}$
$\text{dx}=-\text{dt}\times\text{e}^\text{x}$
Putting $1 + e^{-x} = t$ and $dx = -e^x\ dt$ in equation $(i),$ we get,
$\text{I}=\int\frac{1}{\text{e}^\text{x}\times\text{t}}\times-\text{e}^\text{x}\text{dt}$
$=-\int\frac{\text{dt}}{\text{t}}$
$=-\log|\text{t}|+\text{C}$
$=-\log|1+\text{e}^{-\text{x}}|+\text{C}$
$\therefore -\log|1+\text{e}^{-\text{x}}|+\text{C}$
View full question & answer→Question 2075 Marks
$\int\frac{\text{x}^2+5\text{x}+2}{\text{x}+2}\text{dx}$
Answer$\int\frac{(\text{x}^2+5\text{x}+2)}{(\text{x}+2)}\text{dx}$
$=\int\frac{\text{x}^2}{\text{x}+2}\text{dx}+5\int\frac{\text{x dx}}{\text{x}+2}+2\int\frac{\text{dx}}{\text{x}+2}$
$=\int\Big(\frac{\text{x}^2-4+4}{\text{x}+2}\Big)\text{dx}+5\int\Big(\frac{\text{x}+2-2}{\text{x}+2}\Big)\text{dx}+2\int\frac{\text{dx}}{\text{x}+2}$
$=\int\frac{(\text{x}-2)(\text{x}+2)}{(\text{x}+2)}\text{dx}+\int\frac{4}{\text{x}+2}\text{dx}+5\int\Big(1-\frac{2}{\text{x}+2}\Big)\text{dx}+2\int\frac{\text{dx}}{\text{x}+2}$
$=\int(\text{x}-2)\text{dx}+4\int\frac{\text{dx}}{\text{x}+2}+5\int\text{dx}-10\int\frac{\text{dx}}{\text{x}+2}+2\int\frac{\text{dx}}{\text{x}+2}$
$=\int(\text{x}-2)\text{dx}-4\int\frac{\text{dx}}{\text{x}+2}+5\int\text{dx}$
$=\Big(\frac{\text{x}^2}{2}-2\text{x}\Big)-4\text{ln|}\text{x}+2|+5\text{x}+\text{C}$
$=\frac{\text{x}^2}{2}+3\text{x}-4\text{ln}|\text{x}+2|+\text{C}$
View full question & answer→Question 2085 Marks
Evaluate the following intregals:
$\int\frac{1}{4\cos^2\text{x}+3\sin^2\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{4\cos^2\text{x}+3\sin^2\text{x}}\ \text{dx}$
Dividing numerator and demnominator by $\cos^2\text{x}$
$=\int\frac{\frac{1}{\cos^2\text{x}}}{4+9\tan^2\text{x}} \text {dx}$
$\text{I}=\int\frac{\sec^2\text{x}}{4+9\tan^2\text{x}}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x}\text{ dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{4+9(\text{t})^2}$
$=\int\frac{\text{dt}}{4+(3\text{t})^2}$
Let $3\text{t}=\text{u}$
$3\text{dt}=\text{du}$
$\text{I}=\frac{1}{3}\int\frac{\text{du}}{(2)^2+(\text{u})^2}$
$=\frac{1}{3}\times\frac{1}{2}\times\tan^{-1}\Big(\frac{\text{u}}{2}\Big)+\text{C}$
$\text{I}=\frac{1}{6}\tan^{-1}\Big(\frac{3\text{t}}{2}\Big)+\text{C}$
$\text{I}=\frac{1}{6}\tan^{-1}\Big(\frac{3\tan\text{x}}{2}\Big)+\text{C}$
View full question & answer→Question 2095 Marks
Evaluate the following intregals:
$\int\frac{2\text{x}+1}{(\text{x}-2)(\text{x}-3)}\ \text{dx}$
AnswerLet $\frac{2\text{x}+1}{(\text{x}-2)(\text{x}-3)}=\frac{\text{A}}{(\text{x}-2)}+\frac{\text{B}}{(\text{x}-3)}$
$\Rightarrow2\text{x}+1=\text{A}(\text{x}-3)+\text{B}(\text{x}-2)$
$=(\text{A}+\text{B})\text{x}+(-3\text{A}-2\text{B})$
equating similar terms, we get
A + B = 2, and -3A - 2B = 1
Thus,
$\text{I}=-5\int\frac{\text{dx}}{\text{x}-2}+7\int\frac{\text{dx}}{\text{x}-3}$
$=-5\log|\text{x}-2|+7\log|\text{x}-3|+\text{C}$
$\text{I}=\log\Big|\frac{(\text{x}-3)^7}{(\text{x}-2)^5}\Big|+\text{C}$
View full question & answer→Question 2105 Marks
Evalute the following integrals:
$\int\frac{\cos4\text{x}-\cos2\text{x}}{\sin4\text{x}-\sin2\text{x}}\text{dx}$
Answer$\int\Big(\frac{\cos4\text{x}-\cos2\text{x}}{\sin4\text{x}-\sin2\text{x}}\Big)\text{dx}$
$=\int\frac{-2\sin\Big(\frac{4\text{x}+2\text{x}}{2}\Big)\sin\Big(\frac{4\text{x}-2\text{x}}{2}\Big)}{2\cos\Big(\frac{4\text{x}+2\text{x}}{2}\Big)\sin\Big(\frac{4\text{x}-2\text{x}}{2}\Big)}\text{dx}$
$\bigg[\because\cos\text{A}-\cos\text{B}=-2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\ \& \\ \sin\text{A}-\sin\text{B}=2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\bigg]$
$=-\int\frac{\sin3\text{x}}{\cos3\text{x}}\text{dx}$
$=-\int\tan3\text{x dx}$
$=\frac{-\text{In}|\sec3\text{x}|}{3}+\text{C}$
$=\frac{1}{3}\text{ln}\big(|\sec3\text{x}|\big)^{-1}+\text{C}$
$=\frac{1}{3}\text{ln}|\cos3\text{x}|+\text{C}$
View full question & answer→Question 2115 Marks
Evaluate the following intregals:
$\int\frac{\cos\text{x}}{(1-\sin\text{x})(2-\sin\text{x})}\ \text{dx}$
AnswerWe have,
$\text{I}=\int\frac{\cos\text{x}\text{ dx}}{(1-\sin\text{x})(2-\sin\text{x})}$
putting $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{(1-\text{t})(2-\text{t})}$
$\therefore\text{I}=\int\frac{\text{dt}}{(1-\text{t})(2-\text{t})}$
Let $\frac{1}{(\text{t}-1)(\text{t}-2)}=\frac{\text{A}}{\text{t}-1}+\frac{ \text{B}}{\text{t}-2}$
$\Rightarrow\frac{1}{(\text{t}-1)(\text{t}-2)}=\frac{\text{A}(\text{t}-2)+\text{B}(\text{t}-1)}{(\text{t}-1)(\text{t}-2)}$
$\Rightarrow1=\text{A}(\text{t}-2)+\text{B}(\text{t}-1)$
Putting t - 1 = 0
⇒ t = 1
$\therefore$ 1 = A (1 - 2) + B × 0
⇒ A = -1
Putting t - 2 = 0
⇒ t = 2
$\therefore$ 1 = A × 0 + B(2 - 1)
⇒ B = 1
$\therefore\text{I}=\int\frac{-\text{dt}}{\text{t}-1}+\int\frac{\text{dt}}{\text{t}-2}$
$=-\log|\text{t}-1|+\log|\text{t}-2|+\text{C}$
$=\log\Big|\frac{\text{t}-2}{\text{t}-1}\Big|+\text{C}$
$=\log\Big|\frac{\sin\text{x}-2}{\sin\text{x}-1}\Big|+\text{C}$
$=\log\Big|\frac{2-\sin\text{x}}{1-\sin\text{x}}\Big|+\text{C}$
View full question & answer→Question 2125 Marks
Evaluate the following intregals:
$\int\frac{1}{5-4\sin\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{5-4\sin\text{x}}\ \text{dx}$
Put $\sin\text{x}=\frac{2\tan^\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\text{I}=\int\frac{1}{5-4\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\ \text{dx}$
$=\int\frac{1+\tan^2\frac{\text{x}}{2}}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)-4\Big(2-\tan\frac{\text{x}}{2}\Big)}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{5+5\tan^2\frac{\text{x}}{2}-8\tan\frac{\text{x}}{2}}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$=\int\frac{2\text{dt}}{5\text{t}^2-8\text{t}+5}$
$=\frac{2}{5}\int\frac{\text{dt}}{\text{t}^2-2\text{t}\Big(\frac{4}{5}\Big)+\Big(\frac{4}{5}\Big)^2-\Big(\frac{4}{5}\Big)^2+1}$
$\text{I}=\frac{2}{5}\int\frac{\text{dt}}{\Big(\text{t}-\frac{4}{5}\Big)^2+\Big(\frac{3}{5}\Big)^2}$
$=\frac{2}{5}\times\frac{1}{\frac{3}{5}}\Bigg(\frac{\text{t}-\frac{4}{5}}{\frac{3}{5}}\Bigg)+\text{C}$
$=\frac{2}{3}\tan^{-1}\Big(\frac{5\text{t}-4}{3}\Big)+\text{C}$
$\text{I}=\frac{2}{3}\tan^{-1}\Big(\frac{5\tan\frac{\text{x}}{2}-4}{3}\Big)+\text{C}$
View full question & answer→Question 2135 Marks
Evaluate the following integrals:
$\int\text{e}^{\text{x}}\frac{\text{x}-1}{(\text{x}+1)^3}\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\frac{\text{x}+1-2}{(\text{x}+1)^3}\text{dx}$
$=\int\text{e}^{\text{x}}\Big\{\frac{1}{(\text{x}+1)^2}+\frac{-2}{({\text{x}+1})^3}\Big\}\text{dx}$
$=\int\text{e}^{\text{x}}\frac{1}{(\text{x}+1)^2}\text{dx}+\int\text{e}^{\text{x}}\frac{(-2)}{(\text{x}+1)^3}\text{dx}$
Integrating by parts
$=\text{e}^{\text{x}}\frac{1}{(\text{x}+1)^2}-\int\text{e}^{\text{x}}\Big(\frac{\text{d}}{\text{dx}}(\text{x}+1)^{-2}\Big)\text{dx}+\int\text{e}^{\text{x}}\frac{(-2)}{(\text{x}+1)^3}\text{dx}$
$=\text{e}^{\text{x}}\frac{1}{(\text{x}+1)^2}-\int\text{e}^{\text{x}}\frac{(-2)}{\text{(x+1)}^3}\text{dx}+\int\text{e}^{\text{x}}\frac{(-2)}{(\text{x}+1)^3}\text{dx}$
$=\text{e}^{\text{x}}\frac{1}{(\text{x}+1)}+\text{C}$
View full question & answer→Question 2145 Marks
Evaluate the following intergrals:
$\int\text{e}^\text{ax}\cos\text{bx dx}$
AnswerLet $\text{I}=\int\text{e}^\text{ax}\cos\text{bx dx}$
Intergrating by parts,
$\text{I}=\text{e}^\text{ax}\frac{\sin\text{bx}}{\text{b}}-\text{a}\int\text{e}^\text{ax}\frac{\sin\text{bx}}{\text{x}}\text{dx}$
$=\frac{1}{\text{b}}\text{e}^\text{ax}\sin\text{bx}-\frac{\text{a}}{\text{b}}\int\text{e}^\text{ax}\sin\text{bx dx}$
$=\frac{1}{\text{a}}\text{e}^\text{ax}\sin\text{bx}-\frac{\text{a}}{\text{b}}\Big[-\text{e}^\text{ax}\frac{\cos\text{bx}}{\text{b}}+\int\text{ae}^\text{ax}\frac{\cos\text{bx}}{\text{b}}\text{dx}\Big]$
$=\frac{1}{\text{a}}\text{e}^\text{ax}\sin\text{bx}-\frac{\text{a}}{\text{b}^2}\text{e}^\text{ax}\cos\text{bx}-\frac{\text{a}^2}{\text{b}^2}\int\text{e}^\text{ax}\cos\text{bx dx}$
$\Rightarrow\text{I}=\frac{\text{e}^\text{ax}}{\text{b}^2}\big[\text{b}\sin\text{bx}+\text{a}\cos\text{bx}\big]-\frac{\text{a}^2}{\text{b}^2}\text{I}+\text{C}$
$\Rightarrow\text{I}.\Big\{\frac{\text{a}^2+\text{b}^2}{\text{b}^2}\Big\}=\frac{\text{e}^\text{ax}}{\text{b}^2}\big[\text{b}\cos\text{bx}+\text{a}\cos\text{bx}\big]+\text{C}$
Thus,
$\text{I}=\frac{\text{e}^\text{ax}}{\text{a}^2+\text{b}^2}\big[\text{a}\cos\text{bx}+\text{a}\cos\text{bx}\big]+\text{C}$
View full question & answer→Question 2155 Marks
Evaluate the following integrals:
$\int\text{x}\cos^3\text{x}^2\sin\text{x}^2\text{ dx}$
Answer$\int\text{x}\cos^3\text{x}^2\sin\text{x}^2\text{ dx}$
Let $\text{x}^2=\text{t}$
$2\text{x}\text{dx}=\text{dt}$
$\text{x}\text{ dx}=\frac{\text{dt}}{2}$
Now, $\int\text{x}\cos^3\text{x}^2\sin\text{x}^2\text{ dx}$
$=\frac{1}{2}\int\cos^3\text{t}\cdot\sin\text{t}\text{ dt}$
Again let $\cos\text{t}=\text{p}$
$-\sin\text{t}\text{ dt}=\text{dp}$
$\sin\text{t}\text{ dt}=-\text{dp}$
So, $\frac{1}{2}\int\cos^3\text{t}\cdot\sin\text{t}\text{ dt}$
$=-\frac{1}{2}\text{p}^3\text{ dp}$
$=-\frac{1}{2}\Big(\frac{\text{p}^4}{4}\Big)+\text{C}$
$=-\frac{\text{p}^4}{8}+\text{C}$
$=-\frac{\cos^4\text{t}}{8}+\text{C}$
$=-\frac{\cos^4\text{x}^2}{8}+\text{C}$
View full question & answer→Question 2165 Marks
Evaluate the following intregals:
$\int\frac{2\text{x}+1}{(\text{x}+2)(\text{x}-3)^2}\text{ dx}$
AnswerLet $\frac{2\text{x}+1}{(\text{x}+2)(\text{x}-3)^2}=\frac{\text{A}}{\text{x}+2}+\frac{\text{B}}{\text{x}-3}+\frac{\text{C}}{(\text{x}-3)^2}$
$\Rightarrow 2x + 1 = A (x - 3)^2 + B (x + 2) (x - 3) + C (x + 2)$
$= (A + B)x^2 + (-6A - B + C)x + (9A - 6B + 2C)$
Equating similar terms, we get, $A + B = 0 $
$\Rightarrow A = -B -6A - B +C = 2 $
$\Rightarrow 5B + C = 2 $
$9A - 6B + 2C = 1 $
$\Rightarrow -15B + 2C = 1$
Solving, we get, $\text{B}=\frac{3}{25},\text{C}=\frac{7}{5},\text{A}=-\frac{3}{25}$thus,
$\text{I}=-\frac{3}{25}\int\frac{\text{dx}}{\text{x}+2}+\frac{3}{25}\int\frac{\text{dx}}{\text{x}-3}+\frac{7}{5}\int\frac{\text{dx}}{(\text{x}-3)^2}$
$\text{I}=-\frac{3}{25}\log|\text{x}+2|+\frac{3}{25}\log|\text{x}-3|-\frac{7}{5(\text{x}-3)}+\text{C}$
View full question & answer→Question 2175 Marks
Evalute the following integrals:
$\int\frac{\sin(\text{x}-\text{a})}{\sin(\text{x}-\text{b})}\text{dx}$
AnswerLet $\text{I}=\int\frac{\sin(\text{x}-\text{a})}{\sin(\text{x}-\text{b})}\text{dx}$
Putting x - b = t
⇒ x = b + t
& dx = dt
$\therefore\text{I}=\int\frac{\sin(\text{b}+\text{t}-\text{a})}{\sin\text{t}}\text{dt}$
$=\int\frac{\sin\big\{(\text{b}-\text{a})+\text{t}\big\}}{\sin\text{t}}\text{dt}$
$=\int\frac{\sin(\text{b}-\text{a})\cos\text{t}}{\sin\text{t}}+\int\frac{\cos(\text{b}-\text{a})\sin\text{t}}{\sin\text{t}}\text{dt}$
$=\int\sin(\text{b}-\text{a})\cot\text{t dt}+\int\cos(\text{b}-\text{a})\text{dt}$
$=\sin(\text{b}-\text{a})\text{in}|\sin\text{t}\big|+\text{t}\cos(\text{b}-\text{a})+\text{C}$
$=\sin(\text{b}-\text{a})\text{in}|\sin(\text{x}-\text{b})|+(\text{x}-\text{b})\cos(\text{b}-\text{a})+\text{C}$
$\big[\because\text{t}=\text{x}-\text{b}\big]$
View full question & answer→Question 2185 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2}{(\text{a}-\text{x}^2)^{\frac{3}{2}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2}{(\text{a}-\text{x}^2)^{\frac{3}{2}}}\text{ dx}$
Let $\text{x}=\text{a}\cos\theta$
On differentiating both sides, we get
$\text{dx}=-\text{a}\sin\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\frac{\text{a}^2\cos^2\theta}{(\text{a}^2-\text{a}^2\cos^2\theta)^\frac{3}{2}}\times-\text{a}\sin\theta\text{ d}\theta$
$=-\int\frac{\text{a}^3\cos^2\theta\sin\theta}{\text{a}^3(1-\cos^2\theta)^{\frac{3}{2}}}\text{ d}\theta$
$=-\int\frac{\cos^2\theta\sin\theta}{\sin^3\theta}\text{ d}\theta$
$=-\int\cot^2\theta\text{ d}\theta$
$=-\int(\text{cosec}^2\theta-1)\text{d}\theta$
$=-(-\cot\theta-\theta)+\text{C}$
$=\cot\theta+\theta+\text{C}$
$=\cot\Big(\cos^{-1}\frac{\text{x}}{\text{a}}\Big)+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
$=\cot\Big(\cos^{-1}\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}\Big)+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
$=\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
Hence, $\int\frac{\text{x}^2}{(\text{a}^2-\text{x}^2)^{\frac{3}{2}}}\text{ dx}=\frac{\text{x}}{\sqrt{\text{a}^2-\text{x}^2}}+\cos^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
View full question & answer→Question 2195 Marks
Evaluate the following integrals:
$\int\frac{1}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}\text{ dx}$
AnswerWe have
$\text{I}=\int\frac{1}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}\text{ dx}$
Putting $\text{x}=\frac{1}{\text{t}}$
$\text{dx}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=\int\frac{-\frac{1}{\text{t}^2}\text{dt}}{\big(1+\frac{1}{\text{t}^2}\big)\sqrt{1-\frac{1}{\text{t}^2}}}$
$=\int\frac{-\frac{1}{\text{t}^2}\text{ dt}}{\frac{(\text{t}^2+1)}{\text{t}^2}\frac{\sqrt{\text{t}^2-1}}{\text{t}}}$
$=-\int\frac{\text{t dt}}{(\text{t}^2+1)\sqrt{\text{t}^2-1}}$
Again Putting $\text{t}^2-1=\text{u}^2$
$2\text{tdt}=2\text{udu}$
$\text{tdt}=\text{udu}$
$\therefore\ \text{I}=-\int\frac{\text{u du}}{(\text{u}^2+2)\text{u}}$
$=-\int\frac{\text{du}}{\text{u}^2+(\sqrt{2})^2}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{u}}{\sqrt{2}}\Big)+\text{C}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\bigg(\frac{\sqrt{\text{t}^2-1}}{\sqrt{2}}\bigg)+\text{C}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\Bigg(\sqrt{\frac{\frac{1}{\text{x}^2}-1}{2}}\Bigg)+\text{C}$
$=-\frac{1}{\sqrt{2}}\tan^{-1}\bigg(\sqrt{\frac{1-\text{x}^2}{2\text{x}^2}}\bigg)+\text{C}$
View full question & answer→Question 2205 Marks
Evaluate the following integrals:
$\int\cot^5\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\cot^5\text{x}\text{ dx}$ Then
$\text{I}=\int\cot^3\text{x}\times\big(\text{cosec}^2-1\big)\text{dx}$
$=\int\cot^3\text{x}\times\big(\text{cosec}^2\text{x}-1\big)\text{dx}$
$=\int\cot^3\text{x}\text{ cosec}^2\text{x}\text{ dx}-\int\cot^3\text{x}\text{dx}$
$=\int\cot^3\text{x}\text{ cosec}^2\text{x}\text{ dx}-\int\big(\text{cosec}^2\text{x}-1\big)\cot\text{dx}$
$=\int\cot^3\text{x}\text{cosec}^2\text{x}\text{ dx}-\int\text{cosec}^2\text{x}\cot\text{x}\text{ dx}+\int\cot\text{x}\text{ dx}$
$\text{I}=\int\cot^3\text{x}\text{cosec}^2\text{x}\text{ dx}-\int\text{cosec}^2\text{x}\cot\text{x}\text{ dx}+\int\cot\text{x}\text{ dx}$
Substituting $\cot\text{x}=\text{t}$ and $-\text{cosec}^2\text{x}\text{ dx}=\text{dt}$ in first two integral, we get
$\text{I}=\int\text{t}^3(-\text{dt})-\int\text{t}\times(-\text{dt})+\int\cot\text{x}\text{ dx}$
$=-\frac{\text{t}^4}{4}+\frac{\text{t}^2}{2}+\log|\sin\text{x}|+\text{C}$
$=-\frac{1}{4}\cot^4\text{x}+\frac{1}{2}\cot^2\text{x}+\log|\sin\text{x}|+\text{C}$
$\therefore\ \text{I}=-\frac{1}{4}\cot^4\text{x}+\frac{1}{2}\cot^2\text{x}+\log|\sin\text{x}|+\text{C}$
View full question & answer→Question 2215 Marks
Evaluate the following intregals:
$\int\frac{1}{4\cos\text{x}-1}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{4\cos\text{x}-1}\ \text{dx}$
Putting $\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1-\tan^2\frac{\text{x}}{2}}$
$\Rightarrow\text{I}=\int\frac{1}{4\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)-1}\ \text{dx}$
$=\int\frac{1}{\frac{4\Big(1-\tan^2\frac{\text{x}}{2}\Big)-\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}}$
$=\int\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)\text{dx}}{4-4\tan^2\big(\frac{\text{x}}{2}\big)-1-\tan^2\big(\frac{\text{x}}{2}\big)}$
$=\int\frac{\sec^2\big(\frac{\text{x}}{2}\big)\text{dx}}{3-5\tan^2\big(\frac{\text{x}}{2}\big)}$
Let $\tan\Big(\frac{\text{x}}{2}\Big)=\text{t}$
$\Rightarrow\frac{1}{2}\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=\text{dt}$
$\Rightarrow\sec^2\Big(\frac{\text{x}}{2}\Big)\text{dx}=2\text{dt}$
$\therefore\ \text{I}=2\int\frac{\text{x}}{3-5\text{t}^2}$
$=\frac{2}{5}\int\frac{\text{dt}}{\frac{3}{5}-\text{t}^2}$
$=\frac{2}{5}\int\frac{\text{dt}}{\Big(\frac{\sqrt{3}}{\sqrt{5}}\Big)^2-\text{t}^2}$
$=\frac{2}{5}\times\frac{\sqrt{5}}{2\sqrt{3 }}\ln\begin{vmatrix}\frac{\frac{\sqrt{3}}{\sqrt{5}}+\text{t}}{\frac{\sqrt{3}}{\sqrt{5}}-\text{t}}\end{vmatrix}+\text{C}$
$=\frac{1}{\sqrt{15}}\ln\bigg|\frac{\sqrt{3}+\sqrt{5}\text{t}}{\sqrt{3}-\sqrt{5}\text{t}}\bigg|+\text{C}$
$=\frac{1}{\sqrt{15}}\ln\begin{vmatrix}\frac{\sqrt{3}+\sqrt{5}\tan\big(\frac{\text{x}}{2}\big)}{\sqrt{3}-\sqrt{5}\tan\big(\frac{\text{x}}{2}\big)}\end{vmatrix}+\text{C}$
View full question & answer→Question 2225 Marks
Evaluate the following integrals:
$\int\sqrt{2\text{x}^2+3\text{x}+4}\text{dx}$
Answer$\text{I}=\int\sqrt{2\text{x}^2+3\text{x}+4}\text{dx}$
$=\sqrt2\int\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}\text{dx}$
$=\sqrt2\int\sqrt{\text{x}^2+\frac{3}{2}\text{x}+\frac{9}{16}+\frac{23}{16}}\text{dx}$
$=\sqrt2\int\sqrt{\Big(\text{x}+\frac{3}{4}\Big)^2+\Big(\frac{\sqrt{23}}{4}\Big)^2}\text{dx}$
$=\sqrt2\begin{Bmatrix}\frac{\big(\text{x}+\frac{3}{4}\big)}{2}\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}+\frac{23}{32}\\\times\log\bigg|\Big(\text{x}+\frac{3}{4}\Big)+\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}\bigg|+\text{C}\end{Bmatrix}$
$\therefore\ \text{I}=\frac{4\text{x}+3}{8}\sqrt{2\text{x}^2+3\text{x}+4}+\frac{23\sqrt2}{32}\\\times\log\bigg|\Big(\text{x}+\frac{3}{4}\Big)+\sqrt{\text{x}^2+\frac{3}{2}\text{x}+2}\bigg|+\text{C}$
View full question & answer→Question 2235 Marks
Evaluate the following intregals:
$\int\frac{1}{5+4\cos\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{1}{5+4\cos\text{x}}\ \text{dx}$
Put $\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\text{I}=\int\frac{1}{5+4\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\ \text{dx}$
$=\int\frac{1+\tan^2\frac{\text{x}}{2}}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)+4\Big(1-\tan^2\frac{\text{x}}{2}\Big)}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{9+\tan^2\frac{\text{x}}{2}}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$=\int\frac{2\text{dt}}{(3)^2+\text{t}^2}$
$=2\times\frac{1}{3}\tan^{-1}(\text{t})+\text{C}$
$\text{I}=\frac{2}{3}\tan^{-1}\Big(\frac{\tan\frac{\text{x}}{2}}{3}\Big)+\text{C}$
View full question & answer→Question 2245 Marks
Evaluate the following integrals:
$\int\frac{\text{x}}{(\text{x}^2+1)\sqrt{\text{x}}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\text{x}}{(\text{x}^2+1)\sqrt{\text{x}}}\text{ dx}$
Let $\text{x}=\text{t}^2$
$\text{dx}=2\text{t dt}$
$\therefore\ 2\int\frac{\text{t dt}}{(\text{t}^2+1)\text{t}}$
$=2\Big|\frac{\text{dt}}{\text{t}^4+1}\Big|$
Dividing numerator and denominator by $t^2$
$\text{I}=2\int\frac{\frac{\text{t}}{\text{t}^2}}{\big(\text{t}^2+\frac{1}{\text{t}^2}\big)}\text{ dt}$
$=\int\frac{\Big(1+\frac{1}{\text{t}^2}\big)-\Big(1-\frac{1}{\text{t}^2}\Big)}{\Big(\text{t}^2+\frac{1}{\text{t}^2}\Big)}\text{ dt}$
Let $\text{t}-\frac{1}{\text{t}}=\text{z}$
$\Big(1+\frac{1}{\text{t}^2}\Big)\text{ dt}=\text{dz} [$For $I^{st}$ part$]$
and, $\text{t}+\frac{1}{\text{t}}=\text{y}$
$\Big(1+\frac{1}{\text{t}^2}\Big)\text{ dt}=\text{dy} [$For $II^{nd}$ part$]$
$\therefore\ \text{I}=\int\frac{\text{dz}}{\text{z}^2+2}-\int\frac{\text{dy}}{\text{y}^2-2}$
$=\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{z}}{\sqrt{2}}\Big)-\frac{1}{2\sqrt{2}}\log\bigg|\frac{\text{y}-\sqrt{2}}{\text{y}+\sqrt{2}}\bigg|+\text{C}$
$=\frac{1}{\sqrt{2}}\tan^{1}\Big(\frac{\text{t}^2-1}{\sqrt{2}\text{t}}\Big)-\frac{1}{2\sqrt{2}}\log\bigg|\frac{\text{x}+1-\sqrt{2\text{x}}}{\text{x}+1+\sqrt{2\text{x}}}\bigg|+\text{C}$
View full question & answer→Question 2255 Marks
Evaluate the following integral:
$\int\frac{\text{x}^4+1}{\text{x}^2+1}\text{ dx}$
Answer$\int\Big(\frac{\text{x}^4+1}{\text{x}^2+1}\Big)\text{ dx}$
$=\int\Big(\frac{\text{x}^4-1+1+1}{\text{x}^2+1}\Big)\text{ dx}$
$=\int\Big[\frac{(\text{x}^4-1)}{\text{x}^2+1}+\frac{2}{\text{x}^2+1}\Big]\text{ dx}$
$=\int\Big[\frac{(\text{x}^2-1)(\text{x}^2+1)}{(\text{x}^2+1)}+\frac{2}{\text{x}^2+1}\Big]\text{ dx}$
$=\int\Big[(\text{x}^2-1)+\frac{2}{\text{x}^2+1}\Big]\text{ dx}$
$=\frac{\text{x}^3}{3}-\text{x}+2\tan^{-1}(\text{x})+\text{C}$
View full question & answer→Question 2265 Marks
Evaluate the following intregals:
$\int\frac{\sin2\text{x}}{(1+\sin\text{x})(2+\sin\text{x})}\text{ dx}$
AnswerLet $\int\frac{\sin2\text{x}}{(1+\sin\text{x})(2+\sin\text{x})}\text{ dx}=\frac{\text{A}}{1+\sin\text{x}}+\frac{\text{B}}{2+\sin\text{x}}$
$\Rightarrow\sin2\text{x}=\text{A}(2+\sin\text{B})+\text{B}(1+\sin\text{B})$
$\Rightarrow2\sin\text{x}\cos\text{x}=(2\text{A}+\text{B})+(\text{A}+\text{B})\sin\text{x}$
Equating similar terms, we get,
$2\text{A}+\text{B}=0\Rightarrow\text{B}=-2\text{A}\text{ and}$
$\text{A}+\text{B}=2\cos\Rightarrow-\text{A}=2\cos\text{x}$
$\Rightarrow\text{A}=-2\cos\text{x}$
Thus,
$\text{I}=\int-\frac{2\cos\text{x}}{1+\sin\text{x}}\text{ dx}+\int\frac{4\cos\text{x}}{1+\sin\text{x}}\text{ dx}$
$=-2\log|1+\sin\text{x}|+4\log|2+\sin\text{x}|+\text{C}$
$\text{I}=\log\Big|\frac{(2+\sin\text{x})^4}{(1+\sin\text{x})^2}\Big|+\text{C}$
View full question & answer→Question 2275 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{\text{x}}(\text{x}-4)}{(\text{x}-2)^3}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{\text{x}}(\text{x}-4)}{(\text{x}-2)^3}\text{dx}$
$=\int\text{e}^{\text{x}}\bigg\{\frac{(\text{x}-2)-2}{(\text{x}-2)^3}\bigg\}\text{dx}$
$=\int\text{e}^{\text{x}}\bigg\{\frac{1}{(\text{x}-2)^2}-\frac{2}{(\text{x}-2)^3}\bigg\}\text{dx}$
Here, $\text{f(x)}=\frac{1}{(\text{x}-2)^2}$ and $\text{f}'\text{(x)}=\frac{-2}{(\text{x}-2)^3}$
And we know that,
$\int\text{e}^{\text{ax}}(\text{af(x)}+\text{f}'(\text{x}))\text{dx}=\text{e}^{\text{ax}}\text{f(x)}+\text{C}$
$\therefore\int\text{e}^{\text{x}}\bigg\{\frac{1}{(\text{x}-2)^2}-\frac{2}{(\text{x}-2)^3}\bigg\}\text{dx}=\frac{\text{e}^{\text{x}}}{(\text{x}-2)^2}+\text{C}$
$\therefore\text{I}=\frac{\text{e}^{\text{x}}}{(\text{x}-2)^2}+\text{C}$
View full question & answer→Question 2285 Marks
Evaluate the following integrals:
$\int\sqrt{2\text{x}-\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{2\text{x}-\text{x}^2}\text{dx}$
$=\int\sqrt{\text{x}(2-\text{x})}\text{dx}$
Let $\text{x}=1+\sin\text{u}$
or, $\text{dx}=\cos\text{u du}$
$\Rightarrow\text{I}=\int\sqrt{(1+\sin\text{u})(1-\sin\text{u})}\cos\text{u du}$
$\Rightarrow\text{I}=\int\cos^2\text{u du}$
$\Rightarrow\text{I}=\frac{1}{2}\int(\cos2\text{u}+1)\text{du}$
$\Rightarrow\text{I}=\frac{1}{2}\Big(\frac{1}{2}\sin2\text{u}+\text{u}\Big)+\text{C}$
$\Rightarrow\text{I}=\frac{1}{2}(\sin\text{u}\cos\text{u}+\text{u})+\text{C}$
$\Rightarrow\text{I}=\frac{1}{2}\big(\sin\text{u}\sqrt{1-\sin^2\text{u}}+\text{u}\big)+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}(\text{x}-1)\sqrt{2\text{x}-\text{x}^2}+\frac{1}{2}\sin^{-1}(\text{x}-1)+\text{C}$ $\big[\because\text{u}=\sin^{-1}(\text{x}-1)\big]$
View full question & answer→Question 2295 Marks
Evaluate the following integrals:
$\int\frac{\text{x}^2\sin^{-1}\text{x}}{(1-\text{x}^2)^{\frac{3}{2}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^2\sin^{-1}\text{x dx}}{(1-\text{x}^2)^{\frac{3}{2}}}$
Putting $\text{x}=\sin\theta$
$\Rightarrow\text{dx}=\cos\theta\text{d}\theta$
$\&\ \theta=\sin^{-1}\text{x}$
$\therefore\text{I}=\int\frac{\sin^2\theta.\theta.\cos\theta\text{d}\theta}{(1-\sin^2\theta)^{\frac{3}{2}}}$
$=\int\frac{\sin^2\theta.\theta.\cos\theta\text{d}\theta}{(\cos^2\theta)^{\frac{3}{2}}}$
$=\int\frac{\sin^2\theta.\theta.\cos\theta\text{d}\theta}{\cos^3\theta}$
$=\int\tan^2\theta.\theta.\text{d}\theta$
$=\int(\sec^2\theta-1)\theta.\text{d}\theta$.
$=\int\theta.\sec^2\theta\text{d}\theta-\int\theta.\text{d}\theta$
$=\theta\int\sec^2\theta\text{d}\theta-\int\Big\{\frac{\text{d}}{\text{d}\theta}(\theta)\int\sec^2\theta\text{d}\theta\Big\}\text{d}\theta-\int\theta.\text{d}\theta$
$=\theta\tan\theta-\int1.\tan\theta\text{d}\theta-\frac{\theta^{\ 2}}{2}$
$=\theta.\tan\theta-\ln\big|\sec\theta\big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\theta.\frac{\sin\theta}{\cos\theta}+\ln\big|\cos\theta\big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\theta.\frac{\sin\theta}{\cos\theta}+\ln\Big|\sqrt{1-\sin^2\theta}\Big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\frac{\theta.\sin\theta}{\sqrt{1-\sin^2\theta}}+\frac{1}2\ln\Big|1-\sin^2\theta\Big|-\frac{\theta^{\ 2}}{2}+\text{C}$
$=\frac{\text{x}\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}+\frac{1}{2}\ln\big(1-\text{x}^2\big)-\frac{1}{2}\big(\sin^{-1}\text{x}\big)^2+\text{C}$$\Big[\because\theta=\sin^{-1}\text{x}\Big]$
View full question & answer→Question 2305 Marks
Evaluate the following integrals:
$\int\text{x}\sin^3\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\sin^3\text{x dx}$
$\sin(3\text{A})=3\sin\text{A}-4\sin^3\text{A}$
$\sin^3\text{A}=\frac{1}4{}\big[3\sin\text{A}-\sin3\text{A}\big]$
$\therefore\text{I}=\frac{1}{4}\int\text{x}(3\sin\text{x}-\sin3\text{x})\text{dx}$
$=\frac{3}{4}\int\text{x}\sin\text{x dx}-\frac{1}{4}\int\text{x}\sin(3\text{x})\text{dx}$
$=\frac{3}{4}\big[\text{x}(-\cos\text{x})-\int1.(-\cos\text{x})\text{dx}\big]\\-\frac{1}{4}\Big[\text{x}\Big(-\frac{\cos3\text{x}}{3}\Big)-\int1.\Big(-\frac{\cos3\text{x}}{3}\Big)\text{dx}\Big]$
$=-\frac{3\text{x}\cos\text{x}}{4}+\frac{3}{4}\sin\text{x}+\frac{\text{x}\cos3\text{x}}{12}-\frac{1}{36}\sin3\text{x}+\text{C}$
View full question & answer→Question 2315 Marks
Evaluate the following integrals:
$\int\sqrt{3-2\text{x}-2\text{x}^2}\text{dx}$
Answer$\text{I}=\int\sqrt{3-2\text{x}-2\text{x}^2}\text{dx}$
$=\sqrt2\int\sqrt{\frac{3}{2}-\text{x}-\text{x}^2}\text{dx}$
$=\sqrt2\int\sqrt{\frac{7}{4}-\Big(\frac{1}{4}+\text{x}+\text{x}^2\Big)}\text{dx}$ $\Big[\text{Adding and subtracting }\frac{1}{4}\Big]$
$=\sqrt2\int\sqrt{\Big(\frac{\sqrt7}{2}\Big)^2-\Big(\text{x}+\frac{1}{2}\Big)^2}\text{dx}$
$=\sqrt2\begin{Bmatrix}\frac{\text{x}+\frac{1}{2}}{2}\sqrt{\frac{3}{2}-\text{x}-\text{x}^2}+\frac{7}{8}\sin^{-1}\bigg(\frac{\text{x}+\frac{1}{2}}{\frac{\sqrt7}{2}}\bigg)+\text{C}\end{Bmatrix}$
$\therefore\ \text{I}=\frac{2\text{x}+1}{4}\sqrt{3-2\text{x}-2\text{x}^2}+\frac{7\sqrt2}{8}\sin^{-1}\Big(\frac{2\text{x}+1}{\sqrt7}\Big)+\text{C}$
View full question & answer→Question 2325 Marks
Evaluate the following integrals:
$\int\frac{1}{(2\text{x}^2+3)\sqrt{\text{x}^2-4}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1}{(2\text{x}^2+3)\sqrt{\text{x}^2-4}}\text{ dx}$
Let $\text{x}=\frac{1}{\text{t}}$
$\text{dx}=-\frac{1}{\text{t}^2}\text{ dt}$
$\therefore\ \text{I}=\int\frac{-\frac{1}{\text{t}^2}\text{ dt}}{\Big(\frac{2}{\text{t}^2+3}\Big)\sqrt{\big(\frac{1}{\text{t}^2}-4\big)}}$
$=-\int\frac{\text{t dt}}{(2+3\text{t}^2)\sqrt{1-4\text{t}^2}}$
Let $1-4\text{t}^2=\text{u}^2$
$-8\text{tdt}=2\text{udu}$
$\therefore\ \text{I}=\frac{1}{4}\int\frac{\text{u du}}{\frac{(11-3\text{u})^2}{4}\text{u}}$
$=\frac{1}{3}\int\frac{\text{du}}{\frac{11}{3}-\text{u}^2}$
$=\frac{1}{2\sqrt{33}}\log\begin{vmatrix}\frac{\text{u}-\sqrt{\frac{11}{3}}}{\text{u}+\sqrt{\frac{11}{3}}}+\text{C}\end{vmatrix}$
$=\frac{1}{2\sqrt{33}}\log\begin{vmatrix}\frac{\sqrt{1-4\text{t}^2}-\sqrt{\frac{11}{3}}}{\sqrt{1-4\text{t}^2}+\frac{11}{3}}+\text{C}\end{vmatrix}$
Hence,
$\text{I}=\frac{1}{2\sqrt{33}}\log\bigg|\frac{\sqrt{11}\text{x}+\sqrt{3\text{x}^2-12}}{\sqrt{11}\text{x}-\sqrt{3\text{x}^2-12}}\bigg|+\text{C}$
View full question & answer→Question 2335 Marks
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}(-\sin\text{x}+2\cos\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}}(-\sin\text{x}+2\cos\text{x})\text{dx}$
$=-\int\text{e}^{2\text{x}}\sin\text{x dx}+2\int\text{e}^{2\text{x}}\cos\text{x dx}$
Applying by parts in the $2^{nd}$ integrand
$\therefore\text{I}=-\int\text{e}^{2\text{x}}\sin\text{x dx}+2\Big\{\frac{1}{2}\text{e}^{2\text{x}}\cos\text{x}+\int\frac{1}{2}\text{e}^{2\text{x}}\sin\text{x dx}\Big\}$
$=-\int\text{e}^{2\text{x}}\sin\text{x dx}+\text{e}^{2\text{x}}\cos\text{x}+\int\text{e}^{2\text{x}}\sin\text{x dx}+\text{C}$
$=\text{e}^{2\text{x}}\cos\text{x}+\text{C}$
Thus,
$\text{I}=\text{e}^{2\text{x}}\cos\text{x + C}$
View full question & answer→Question 2345 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^2-10\text{x}+34}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{\text{x}^2-10\text{x}+34}\text{dx}$
$=\int\frac{1}{\text{x}^2-2\text{x}\times5+(5)^2-(5)^2+34}\text{dx}$
$=\int\frac{1}{(\text{x}-5)^2+9}\text{dx}$
Let $(\text{x}-1)=\text{t} \dots(1)$
$\Rightarrow\text{dx = dt}$
so,
$\text{I}=\int\frac{1}{\text{t}^2+(3)^2}\text{dt}$
$\text{I}=\frac{1}{3}\tan^{-1}\big(\frac{\text{t}}{3}\big)+\text{C}$ $\Big[\text{since,}\int\frac{1}{\text{x}^2+\text{a}^2}\text{dx}=\frac{1}{\text{a}}\tan^{-1}\big(\frac{\text{x}}{2}\big)+\text{C}\Big]$
$\text{I}=\frac{1}{3}\tan^{-1}\Big(\frac{\text{x}-5}{3}\Big)+\text{C}$ [using (1)]
View full question & answer→Question 2355 Marks
Evaluate the following integrals: $\int\cos(\log\text{x})\text{dx}$
AnswerLet $\text{I}=\int\cos(\log\text{x})\text{dx}$
Let $\log\text{x}=\text{t}$
$\Rightarrow\text{x}=\text{e}^\text{t}$
$\Rightarrow\text{dx}=\text{e}^\text{t}\text{dt}$
$\text{I}=\int\text{e}^\text{t}\cos\text{(t)dt}$
Considering $\cos (t)$ as first function and $e^t$ as second function
$\text{I}=\cos\text{t}\text{e}^\text{t}-\int(-\sin\text{t})\text{e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\cos\text{t}\text{e}^\text{t}+\int\sin\text{t}\text{e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\cos\text{te}^\text{t}+\text{I}_1\ \dots(1)$
where $\text{I}_1=\int\text{e}^\text{t}\sin\text{t dt}$
$\text{I}_1=\int\text{e}^\text{t}\sin\text{t dt}$
Considering $\sin t$ as first function and $e^t$ as second function
$\text{I}_1=\sin\text{te}^\text{t}-\int\cos\text{te}^\text{t}\text{dt}$
$\Rightarrow\text{I}_1=\sin\text{te}^\text{t}-\text{I}\ \dots(2)$
From $(1) (2)$
$\text{I}=\cos\text{te}^\text{t}+\sin\text{te}^\text{t}-\text{I}$
$\Rightarrow2\text{I}=\text{e}^\text{t}(\sin\text{t}+\cos\text{t})$
$\Rightarrow\text{I}=\frac{\text{e}^\text{t}(\sin\text{t}+\cos\text{t})}{2}+\text{C}$
$\Rightarrow\text{I}=\frac{\text{e}^{\log\text{x}}\big[\sin(\log\text{x})+\cos(\log\text{x})\big]}{2}+\text{C}$
$\Rightarrow\text{I}=\frac{\text{x}}{2}\big[\sin(\log\text{x})+\cos(\log\text{x})\big]+\text{C}$
View full question & answer→Question 2365 Marks
Evaluvate the following intregals:
$\int\frac{2\sin\text{x}+3\cos\text{x}}{3\sin\text{x}+4\cos\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{2\sin\text{x}+3\cos\text{x}}{3\sin\text{x}+4\cos\text{x}}\ \text{dx}$
Let $(2\sin\text{x}+3\cos\text{x})=\lambda\frac{\text{d}}{\text{dx}}(3\sin\text{x}+4\cos\text{x})+\mu(3\sin\text{x}+4\cos\text{x})+\text{v}$
$(2\sin\text{x}+3\cos\text{x})=\lambda(3\cos\text{x}-4\sin\text{x})+\mu(3\sin\text{x}+4\cos\text{x})+\text{v}$
$(2\sin\text{x}+3\cos\text{x})=(3\lambda+4\mu)\cos\text{x}+(-4\lambda+3\mu)\sin\text{x}+\text{v}$
Compairing the coefficient of $\sin\text{x},\cos\text{x}$ on both the sides,
$3\lambda+4\mu=3\dots\dots(1)$
$-4\lambda+3\mu=2\dots\dots(2)$
$\text{v}=0\dots\dots(3)$
Solving the equation (1), (2) and (3)
$\lambda=\frac{1}{25}$
$\mu=\frac{18}{25}$
$\text{v}=0$
$\text{I}=\frac{1}{25}\int\frac{(3\cos\text{x}-4\sin\text{x})}{(3\sin\text{x}+4\cos\text{x})}\text{dx}+\frac{18}{25}\int\text{dx}$
$\text{I}=\frac{1}{25}\log|3\sin\text{x}+4\cos\text{x}|+\frac{18}{25}\text{x}+\text{C}$
View full question & answer→Question 2375 Marks
Evaluate the following integrals:
$\int\frac{\text{x}}{\text{x}^4+2\text{x}^2+3}\text{dx}$
Answer$\int\frac{\text{x dx}}{\text{x}^4+2\text{x}^2+3}$
Let $\text{x}^2=\text{t}$
$\Rightarrow2\text{x dx = dt}$
$\Rightarrow\text{x dx}=\frac{\text{dt}}{2}$
Now, $\int\frac{\text{x dx}}{\text{x}^4+2\text{x}^2+3}$
$=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2+2\text{t}+3}$
$=\frac{1}{2}\int\frac{\text{dt}}{\text{t}^2+2\text{t}+1+2}$
$=\frac{1}{2}\int\frac{\text{dt}}{(\text{t}+1)^2+(\sqrt2)^2}$
$=\frac{1}{2}\times\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{t}+1}{\sqrt{2}}\Big)+\text{C}$ $\Big[\because\int\frac{\text{dx}}{\text{x}^2+\text{a}^2}=\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)+\text{C}\Big]$
$=\frac{1}{2\sqrt{2}}\tan^{}-1\Big(\frac{\text{x}^2+1}{\sqrt{2}}\Big)+\text{C}$
View full question & answer→Question 2385 Marks
Evaluate the following integrals:
$\int\text{cosec}^43\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\text{cosec}^43\text{x}\text{ dx}$ Then
$\text{I}=\int\text{cosec}^23\text{x }\text{cosec}^23\text{x}\text{ dx}$
$=\int\big(1+\cot^23\text{x}\big)\text{cosec}^23\text{x}\text{ dx}$
$=\int\big(\text{cosec}^23\text{x}+\cot^23\text{x }\text{cosec}^23\text{x}\big)\text{dx}$
$\text{I}=\int\text{cosec}^23\text{x}\text{ dx}+\int\cot^23\text{x }\text{cosec}^23\text{x}\text{ dx}$
Sunbstituting $\cot3\text{x}=\text{t}$ and $\text{cosec}^23\text{x}\text{ dx}=-\text{dt}$ in $2^{nd}$ integral, we get
$\text{I}=\int\text{cosec}^23\text{x}-\int\text{t}^2\frac{\text{dt}}{3}$
$=\frac{-1}{3}\cot3\text{x}-\frac{\text{t}^3}{9}+\text{C}$
$=\frac{-1}{3}\cot3\text{x}-\frac{\cot^33\text{x}}{9}+\text{C}$
$\therefore\ \text{I}=\frac{-1}{3}\cot3\text{x}-\frac{1}{9}\infty\text{t}^3\text{3x}+\text{C}$
View full question & answer→Question 2395 Marks
Evaluate the following intregals:
$\int\frac{3+4\text{x}-\text{x}^2}{(\text{x}+2)(\text{x}-1)}\ \text{dx}$
Answer$\text{I}=\int\frac{3+4\text{x}-\text{x}^2}{(\text{x}+2)(\text{x}-1)}\ \text{dx}$
$=\int-1+\frac{5\text{x}+1}{(\text{x}+2)(\text{x}-1)}\ \text{dx}$
$\Rightarrow\text{I}=-\int\text{dx}+\int\frac{5\text{x}+1}{(\text{x}+2)(\text{x}-1)}\ \text{dx}\ \dots(1)$
Let $\frac{5\text{x}+1}{(\text{x}+2)(\text{x}-1)}=\frac{\text{A}}{\text{x}+2}+\frac{\text{B}}{\text{x}-1}$
$\Rightarrow5\text{x}+1=\text{A}(\text{x}-1)+\text{B}(\text{x}+2)$
put x = 1
$\Rightarrow6=3\text{B}\Rightarrow\text{B}=2$
Put x = -2
$\Rightarrow-9=-3\text{A}\Rightarrow\text{A}=3$
So,
$\text{I}=\int\text{dx}+3\int\frac{\text{dx}}{\text{x}+2}+2\int\frac{\text{dx}}{\text{x}-1}$
$\text{I}=-\text{x}+3\log|\text{x}+2|+2\log|\text{x}-1|+\text{C}$
View full question & answer→Question 2405 Marks
Evaluate the following integrals:
$\int\text{x}\sin\text{x}\cos2\text{x dx}$
Answer$\int\text{x}.\cos2\text{x}\sin\text{x dx}$
$=\frac{1}{2}\int\text{x}(2\cos2\text{x}\sin\text{x})\text{dx}$ $\big[\therefore2\cos\text{A}\sin\text{B}=\sin(\text{A+B})-\sin(\text{A}-\text{B})\big]$
$=\frac{1}{2}\int\text{x}(\sin3\text{x}-\sin\text{x})\text{dx}$
$=\frac{1}{2}\int\text{x}\sin3\text{x dx}-\frac{1}{2}\int\text{x}\sin\text{x dx}$
$=\frac{1}{2}\int\text{x}\sin3\text{x dx}-\frac{1}{2}\int\text{x}\sin\text{x dx}$
$=\frac{1}{2}\Big[\text{x}\int\sin3\text{x dx}-\int\Big\{\frac{\text{x}}{\text{dx}}(\text{x})\int\sin3\text{x dx}\Big\}\text{dx}\Big]\\-\frac{1}{2}\Big[\text{x}\int\sin\text{x dx}-\int\Big\{\frac{\text{d}}{\text{dx}}(\text{x})\int\sin\text{x dx}\Big\}\text{dx}\Big]$
$=\frac{1}{2}\Big[\text{x}\Big(\frac{-\cos3\text{x}}{3}\Big)-\int1\Big(\frac{-\cos3\text{x}}{3}\Big)\text{dx}\Big]\\-\frac{1}{2}\big[\text{x}(-\cos\text{x})-\int1(-\cos\text{x})\text{dx}\big]$
$=\frac{1}{2}\Big[\text{x}\Big(\frac{-\cos3\text{x}}{3}\Big)+\frac{1}{9}\sin3\text{x}\big]-\frac{1}{2}\big[\text{x}(-\cos\text{x})+\sin\text{x}\big]$
$=-\frac{\text{x}\cos3\text{x}}{6}+\frac{\sin3\text{x}}{18}+\frac{\text{x}\cos\text{x}}{2}-\frac{\sin\text{x}}{2}+\text{C}$
View full question & answer→Question 2415 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2-1}\text{ dx}\int\frac{\text{x}^2+\text{x}+1}{\text{x}^2-1}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\text{x}^3+\text{x}+1}{\text{x}^2-1}\ \text{dx}$
$=\int\Big(\text{x}+\frac{2\text{x}+1}{\text{x}^2-1}\Big)\text{dx}$
Now,
$\frac{2\text{x}+1}{\text{x}^2-1}=\frac{\text{A}}{\text{x}+1}+\frac{\text{B}}{\text{x}-1}$
$\Rightarrow2\text{x}+1=\text{A}(\text{x}-1)+\text{B}(\text{x}+1)$
put x = 1
$\Rightarrow3=2\text{B}\Rightarrow\text{B}=\frac{3}{2}$
put x = -1
$\Rightarrow-1=-2\text{A}\Rightarrow\text{A}=\frac{1}{2}$
$\therefore\text{I}=\int\text{xdx}+\frac{1}{2}\int\frac{\text{dx}}{\text{x}+1}+\frac{3}{2}\int\frac{\text{dx}}{\text{x}-1}$
$\text{I}=\frac{\text{x}^2}{2}+\frac{1}{2}\log|\text{x}+1|+\frac{3}{2}\log|\text{x}-1|+\text{C}$
View full question & answer→Question 2425 Marks
Evaluate the follwing intregals:
$\int\frac{2\text{x}}{(\text{x}^2+1)(\text{x}^2+2)^2}\ \text{dx}$
Answer$\int\frac{2\text{x}}{(\text{x}^2+1)(\text{x}^2+2)^2}\ \text{dx}$
Let $\text{x} ^2=\text{y}$
$\Rightarrow2\text{x dx}=\text{dy}$
$\Rightarrow\text{dx}=\frac{\text{dy}}{2\text{x}}$
$\int\frac{2\text{x}}{(\text{x}^2+1)(\text{x}^2+2)^2}\ \text{dx}$
$=\int\frac{\text{dy}}{(\text{y}+1)(\text{y}+2)^2}$
Let $\frac{1}{(\text{y}+1)(\text{y}+2)^2}=\frac{\text{A}}{\text{y}+1}+\frac{\text{B}}{\text{y}+2}+\frac{\text{C}}{(\text{y}+2)^2}\ ...(1)$
$\Rightarrow1=\text{A}(\text{y}+2)^2+\text{B}(\text{y}+1)(\text{y}+2)+\text{C}(\text{y}+1)\ ...(2)$
Putting $y = -2$ in $(2)$
$1 = C (-2 + 1)$
$\Rightarrow C = -1$
Putting $y = -1$ in $(2)$
$1 = A (-1 + 2)^2$
$\Rightarrow 1 = A (1)$
$\Rightarrow A = 1$
Putting $y = 0$ in $(2)$
$1 = 4A + B(2) + C$
$\Rightarrow 1 = 4 + 2B - 1$
$\Rightarrow -2 = 2B$
$\Rightarrow B = -1$
Substituting the values of $A, B$ and $C$ in $(1)$
$\frac{1}{(\text{y}+1)(\text{y}+2)^2}=\frac{1}{\text{y}+1}-\frac{1}{\text{y}+2}-\frac{1}{(\text{y}+2)^2}$
$\Rightarrow\int\frac{\text{dy}}{(\text{y}+1)(\text{y}+2)^2}=\int\frac{\text{dy}}{\text{y}+1}-\int\frac{\text{dy}}{\text{y}+2}-\int\frac{\text{dy}}{(\text{y}+2)^2}$
$=\log|\text{y}+1|-\log|\text{y}+2|+\frac{1}{\text{y}+2}+\text{C}$
Hence, $\int\frac{2\text{x}}{(\text{x}^2+1)(\text{x}^2+2)^2}\ \text{dx}=\log|\text{x}^2+1|-\log|\text{x}^2+2|+\frac{1}{\text{x}^2+2}+\text{C}$
View full question & answer→Question 2435 Marks
Evaluate the following integrals:
$\int\cos^3\sqrt{\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\cos^3\sqrt{\text{x}}\text{dx}$
Let $\text{x}=\text{t}^2$
$\text{dx}=2\text{t dt }$
$=2\int\text{t}\cos^3\text{t dt}$
$=2\int\text{t}\Big(\frac{3\cos\text{t}+\cos3\text{t}}{4}\Big)\text{dt}$
$=\frac{1}{2}\int\text{t}(3\cos\text{t}+\cos3\text{t})\text{dt}$
Using integral\tion by parts,
$\text{I}=\frac{1}{2}\Big[\text{t}\Big(3\sin\text{t}+\frac{1}{3}\sin3\text{t}\Big)+\int\Big(1\times3\sin\text{t}+\frac{\sin3\text{t}}{3}\Big)\text{dt}\Big]$
$=\frac{1}{2}\Big[\text{t}\Big(\frac{9\sin\text{t}+\sin3\text{t}}{3}\Big)+3\cos\text{t}+\frac{\cos3\text{t}}{9}\Big]+\text{C}$
$=\frac{1}{18}\big[27\text{t}\sin\text{t}+3\text{t}\sin3\text{t}+9\cos\text{t}+\cos3\text{t}\big]+\text{C}$
$\text{I}=\frac{1}{18}\big[27\sqrt{\text{x}}\sin\sqrt{\text{x}}+3\sqrt{\text{x}}\sin3\sqrt{\text{x}}+9\cos\sqrt{\text{x}}+\cos3\sqrt{\text{x}}\big]+\text{C}$
View full question & answer→Question 2445 Marks
Evaluate the following intregals:
$\int\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}\ \text{dx}$
AnswerConsider the integrals
$\text{I}=\int\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}\ \text{dx}$
Let $y = x^2$
Thus,
$\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}=\frac{\text{y}+1}{(\text{y}+4)(\text{y}+25)}$
$\Rightarrow\frac{\text{y}+1}{(\text{y}+4)(\text{y}+25)}=\frac{\text{A}}{\text{y}+4}+\frac{\text{B}}{\text{y}+25}$
$\Rightarrow\frac{\text{y}+1}{(\text{y}+4)(\text{y}+25)}=\frac{\text{A}(\text{y}+25)+\text{B}(\text{y}+4)}{(\text{y}+4)(\text{y}+25)}$
$\Rightarrow\text{y}+1=\text{Ay}+25\text{A}+\text{By}+4\text{B}$
Compairing the coefficient, we have,
$A + B = 1$ and $25A + 4B = 1$
Solving the above equations, we have,
$\text{A}=\frac{-1}{7}\text{ and }\text{B}=\frac{8}{7}$
Thus, $\int\frac{\text{x}^2+1}{(\text{x}^2+4)(\text{x}^2+25)}\ \text{dx}$
$=\int\frac{\frac{1}{7}}{\text{x}^2+4}\ \text{dx}+\int\frac{\frac{8}{7}}{\text{x}^2+25}\ \text{dx}$
$=\frac{-1}{7}\int\frac{1}{\text{x}^2+4}\text{ dx}+\frac{8}{7}\int\frac{1}{\text{x}^2+25}\ \text{dx}$
$=\frac{-1}{7}\times\frac{1}{2}\tan^{-1}\frac{\text{x}}{2}+\frac{8}{7}\times\frac{1}{5}\tan^{-1}\frac{\text{x}}{5}+\text{C}$
$=\frac{-1}{14}\tan^{-1}\frac{\text{x}}{2}+\frac{8}{35}\tan^{-1}\frac{\text{x}}{5}+\text{C}$
View full question & answer→Question 2455 Marks
$\int(\text{x}+2)\sqrt{3\text{x}+5}\text{ dx}$
Answer$\text{Let I}=\int(\text{x}+2)\sqrt{3\text{x}+5}\text{ dx}$
$\text{Putting}\ 3\text{x}+5=\text{t}$
$\Rightarrow\text{x}=\frac{\text{t}-5}{3}$
$\Rightarrow3\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{3}$
$\therefore\text{I}=\int\Big(\frac{\text{t}-5}{3}+2\Big)\sqrt{\text{t}}\frac{\text{dt}}{3}$
$=\frac{1}{3}\int\Big(\frac{\text{t}-5+6}{3}\Big)\sqrt{t}\text{ dt}$
$=\frac{1}{9}\int(\text{t}+1)\sqrt{\text{t}}\text{ dt}$
$=\frac{1}{9}\int\Big(\text{t}^\frac{3}{2}+\text{t}^\frac{1}{2}\Big)\text{dt}$
$=\frac{1}{9}\bigg[\frac{2}{5}(3\text{x}+5)^\frac{5}{2}+\frac{2}{3}(3\text{x}+5)^\frac{3}{2}\bigg]+\text{C}$ $[\because\text{t}=3\text{x}+5]$
$=\frac{2}{9}\bigg[(3\text{x}+5)^\frac{3}{2}\Big\{\frac{3\text{x}+5}{5}+\frac{1}{3}\Big\}\bigg]+\text{C}$
$=\frac{2}{9}\bigg[(3\text{x}+5)^\frac{3}{2}\Big\{\frac{9\text{x}+15+5}{15}\Big\}\bigg]+\text{C}$
$=\frac{2}{9}\bigg[(3\text{x}+5)^\frac{3}{2}\Big\{\frac{9\text{x}+20}{15}\Big\}\bigg]+\text{C}$
$=\frac{2}{135}(3\text{x}+5)^\frac{3}{2}(9\text{x}+20)+\text{C}$
View full question & answer→Question 2465 Marks
Evaluate the following integrals:
$\int\frac{1}{1+\sqrt{\text{x}}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1}{1+\sqrt{\text{x}}}\text{dx}\ \ ....(1)$
Let $\text{x}=\text{t}^2$ then,
$\text{dx}=\text{d}(\text{t}^2)$
$\Rightarrow\text{dx}=2\text{t}\text{ dt}$
Putting $\text{x}=\text{t}^2$ and $\text{dx}=2\text{t}\text{ dt}$ in equation (1), we get
$\text{I}=\int\frac{2\text{t}}{1+\sqrt{\text{t}^2}}\text{dt}$
$=\int\frac{2\text{t}}{1+\text{t}}\text{dt} $
$=2\int\frac{\text{t}}{1+\text{t}}\text{dt}$
$=2\int\frac{1+\text{t}-1}{1+\text{t}}\text{dt}$
$=2\int\Big[\frac{1+\text{t}}{1+\text{t}}-\frac{1}{1+\text{t}}\Big]\text{dt}$
$=2\int\text{dt}-2\int\frac{1}{1+\text{t}}\text{dt} $
$=2\text{t}-2\log(1+\text{t})+\text{C}$
$=2\sqrt{\text{x}}-2\log(1+\sqrt{\text{x}})+\text{C}$
$\therefore\text{ I}=2\sqrt{\text{x}}-2\log(1+\sqrt{\text{x}})+\text{C}$
View full question & answer→Question 2475 Marks
Evaluate the following integrals:
$\int\frac{\text{e}^{\text{x}}}{\text{x}}\Big\{\text{x}(\log\text{x})^2+2\log\text{x}\Big\}\text{dx}$
AnswerWe have,
$\text{I}=\int\frac{\text{e}^{\text{x}}}{\text{x}}\Big\{\text{x}(\log\text{x})^2+2\log\text{x}\Big\}\text{dx}$
$=\int\text{e}^{\text{x}}\Big\{(\log\text{x})^2+\frac{2}{\text{x}}\log\text{x}\Big\}\text{dx}$
$=\int\text{e}^{\text{x}}(\log\text{x})^2+2\int\frac{\text{e}^{\text{x}}}{\text{x}}\log\text{x dx}$
Integrating by parts
$=\text{e}^{\text{x}}(\log\text{x})^2-\int\text{e}^{\text{x}}\frac{\text{d}}{\text{dx}}(\log\text{x})^2\text{dx}+2\int\text{e}^{\text{x}}\frac{1}{\text{x}}\log\text{x dx}$
$=\text{e}^{\text{x}}(\log\text{x})^2-\int\text{e}^{\text{x}}\frac{2\log\text{x}}{\text{x}}\text{dx}+2\int\text{e}^\text{x}\frac{\log\text{x}}{\text{x}}\text{dx}$
$=\text{e}^{\text{x}}(\log\text{x)}^2+\text{C}$
View full question & answer→Question 2485 Marks
Evaluate the following integrals:
$\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{ dx}$
Let $\text{x}=\tan\theta$
On differentiating both sides, we get
$\text{dx}=\sec^2\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\frac{\sqrt{1+\tan^2\theta}}{\tan^4\theta}\sec^2\theta\text{ d}\theta$
$=\int\frac{\sec^{3}\theta}{\tan^4\theta}\text{ d}\theta$
$=\int\frac{\cos\theta}{\sin^4\theta}\text{ d}\theta$
$=\int\cot\theta\text{cosec}^3\theta\text{ d}\theta$
Let $\text{cosec}^3\theta=\text{t}$
On differentiating both sides, we get
$-3\text{cosec}^3\theta\cot\theta\text{ d}\theta=\text{dt}$
$\therefore\ \text{I}=-\frac{1}{3}\int\cot\theta\text{cosec}^3\theta\frac{\text{dt}}{\text{cosec}^3\theta\cot\theta}$
$=-\frac{\text{t}}{3}+\text{C}$
$=-\frac{1}{3}(\text{cosec}^3\theta)+\text{C}$
$=-\frac{1}{3}(\text{cosec}(\tan^{-1}\text{x}))^3+\text{C}$
$=-\frac{1}{3}\bigg(\text{cosec}\Big(\text{cosec}^1\frac{\sqrt{1+\text{x}^2}}{\text{x}}\Big)\bigg)^3+\text{C}$
$=-\frac{1}{3}\bigg(\frac{\sqrt{1+\text{x}^2}}{\text{x}}\bigg)^3+\text{C}$
Hence, $\int\frac{\sqrt{1+\text{x}^2}}{\text{x}^4}\text{ dx}=-\frac{1}{3}\bigg(\frac{\sqrt{1+\text{x}^2}}{\text{x}}\bigg)^3+\text{C}$
View full question & answer→Question 2495 Marks
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\sin\text{x}\cos\text{x }\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{2\text{x}}\sin\text{x}\cos\text{x }\text{dx}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}2\sin\text{x}\cos\text{x dx}$
$=\frac{1}{2}\int\text{e}^{2\text{x}}\sin2\text{x dx}$
We know that
$\int\text{e}^{2\text{x}}\sin\text{bx dx}=\frac{\text{e}^{2\text{x}}}{\text{a}^2+\text{b}^2}\{\text{a}\sin\text{bx}-\text{b}\cos\text{bx}\}+\text{C}$
$\Rightarrow\int\text{e}^{2\text{x}}\sin\text{2x dx}=\frac{\text{e}^{2\text{x}}}{8}\{2\sin2\text{x}-2\cos2\text{x}\}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}.\frac{\text{e}^{2\text{x}}}{8}\{2\sin2\text{x}-2\cos2\text{x}\}+\text{C}$
$\therefore\ \text{I}=\frac{\text{e}^{2\text{x}}}{8}\{\sin2\text{x}-\cos2\text{x}\}+\text{C}$
View full question & answer→Question 2505 Marks
Evaluate the following intregals:
$\int\frac{2}{2+\sin^22\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{2}{2+\sin^22\text{x}}\text{ dx}$
$\text{I}=\int\frac{2}{2+2\sin\text{x}\cos\text{x}}\ \text{dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{\frac{1}{\cos^2\text{x}}}{\frac{1}{\cos^2\text{x}}+\frac{\sin\text{x}\cos\text{x}}{\cos^2\text{x}}}\ \text{dx}$
$=\int\frac{\sec^2\text{x}}{\sec^2\text{x}+\tan\text{x}}\ \text{dx}$
$\text{I}=\int\frac{\sec^2\text{x}}{1+\tan^2\text{x}+\tan\text{x}}\ \text{dx}$
Let $\tan\text{x}=\text{t}$
$\sec^2\text{x dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2+\text{t}+1}$
$=\int\frac{\text{dt}}{\text{t}^2+2\text{t}\Big(\frac{1}{2}\Big)+\Big(\frac{1}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2+1}$
$\text{I}=\int\frac{\text{dt}}{\Big(\text{t}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}$
$=\frac{1}{\frac{\sqrt{3}}{2}}\tan^{-1}\Bigg(\frac{\text{t}+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\Bigg)+\text{C}$
$=\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{2\text{t}+1}{\sqrt{3}}\Big)+\text{C}$
$\text{I}=\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{2\tan\text{x}+1}{\sqrt{3}}\Big)+\text{C}$
View full question & answer→