Question 1513 Marks
Evaluate the following integrals:$\int\frac{\cos2\text{x}}{\sqrt{\sin^22\text{x}+8}}\text{ dx}$
Answer$\int\frac{\cos(2\text{x}).\text{dx}}{\sqrt{\sin^22\text{x}+8}}$
Let $\sin(2\text{x})=\text{t}$
$\Rightarrow\cos(2\text{x})\times2.\text{dx}=\text{dt}$
$\Rightarrow\cos(2\text{x}).\text{dx}=\frac{\text{dt}}{2}$
Now, $\int\frac{\cos(2\text{x}).\text{dx}}{\sqrt{\sin^22\text{x}+8}}$
$=\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}^2+\big(2\sqrt2\big)^2}}$
$=\frac{1}{2}\log\Big|\text{t}+\sqrt{\text{t}^2+8}\Big|+\text{C}$
$=\frac{1}{2}\log\Big|\sin(2\text{x})+\sqrt{\sin^2(2\text{x})+8}\Big|+\text{C}$
View full question & answer→Question 1523 Marks
Evaluate the following integrals:
$\int\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}\text{dx}$
Answer$\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}=\frac{\cos\text{x}-\sin\text{x}}{(\sin^2\text{x}+\cos^2\text{x})+2\sin\text{x}\cos\text{x}}$
$[\sin^2\text{x}+\cos^2\text{x}=1;\ \sin2\text{x}=2\sin\text{x}\cos\text{x}]$
$=\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}$
$\sin\text{x}+\cos\text{x}=\text{t}$
$\therefore(\sin\text{x}+\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow\int\frac{\cos\text{x}-\sin\text{x}}{1+\sin2\text{x}}\text{dx}=\int\frac{\cos\text{x}-\sin\text{x}}{(\sin\text{x}+\cos\text{x})^2}\text{dx}$
$=\int\frac{\text{dt}}{\text{t}^2} $
$=\int\text{t}^{-2}\text{dt}$
$=-\text{t}^{-1}+\text{C}$
$=-\frac{1}{\text{t}}+\text{C}$
$=\frac{-1}{\sin\text{x}+\cos\text{x}}+\text{C}$
View full question & answer→Question 1533 Marks
$\int\frac{1}{1-\sin\frac{\text{x}}{2}}\text{dx}$
Answer$\int\frac{\text{dx}}{1-\sin\big(\frac{\text{x}}{2}\big)}$
$=\int\frac{\big(1+\sin\frac{\text{x}}{2}\big)}{\big(1-\sin\frac{\text{x}}{2}\big)\big(1+\sin\frac{\text{x}}{2}\big)}\text{dx}$
$=\int\bigg(\frac{1+\sin\frac{\text{x}}{2}}{1-\sin^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int\bigg(\frac{1+\sin\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}}\bigg)\text{dx}$
$=\int\big(\sec^2\frac{\text{x}}{2}+\sec\frac{\text{x}}{2}\tan\frac{\text{x}}{2}\big)\text{dx}$
$=\frac{\tan\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}+\frac{\sec\big(\frac{\text{x}}{2}\big)}{\frac{1}{2}}+\text{c}$
$=2\big(\tan\frac{\text{x}}{2}+\sec\frac{\text{x}}{2}\big)+\text{c}$
View full question & answer→Question 1543 Marks
Evalute the following integrals:
$\int\frac{\cot\text{x}}{\log\sin\text{x}}\text{dx}$
AnswerNote: Here we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$
Let $\text{I}=\int\frac{\cot\text{x}}{\log\sin\text{x}}\text{dx}$
Putting $\log\sin\text{x}=\text{t}$
$\Rightarrow\cot\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\cot\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\log|\text{t}|+\text{C}$
$=\log|\log\sin\text{x}|+\text{C}\ \big[\because\text{t}=\log\sin\text{x}\big]$
View full question & answer→Question 1553 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{8+3\text{x}-\text{x}^2}}\text{ dx}$
Answer$8 + 3x - x^2$ can be written as $8-\Big(\text{x}^2-3\text{x}+\frac{9}{4}-\frac{9}{4}\Big).$Therefore,
$8-\Big(\text{x}^2-3\text{x}+\frac{9}{4}-\frac{9}{4}\Big)$
$=\frac{41}{4}-\Big(\text{x}-\frac{3}{2}\Big)^2$
$=\int\frac{1}{8+3\text{x}-\text{x}^2}\text{ dx}$
$=\int\frac{1}{\sqrt{\frac{41}{4}-\big(\text{x}-\frac{3}{2}\big)^2}}\text{ dx}$
Let $\text{x}-\frac{3}{2}=\text{t}$
$\therefore\ \text{dx}=\text{dt}$
$=\int\frac{1}{\sqrt{\frac{41}{4}-\big(\text{x}-\frac{3}{2}\big)^2}}\text{ dx}$
$=\int\frac{1}{\sqrt{\Big(\frac{\sqrt{41}}{2}\Big)^2-\text{t}^2}}\text{ dt}$
$=\sin^{-1}\bigg(\frac{\text{t}}{\frac{\sqrt{41}}{2}}\bigg)+\text{C}$
$=\sin^{-1}\bigg(\frac{\text{x}-\frac{3}{2}}{\frac{\sqrt{41}}{2}}\bigg)+\text{C}$
$=\sin^{-1}\Big(\frac{2\text{x}-3}{\sqrt{41}}\Big)+\text{C}$
View full question & answer→Question 1563 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}^2+\text{x}+1}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{\text{x}^2+\text{x}+1}\text{dx}$
$=\int\sqrt{\text{x}^2+\text{x}+\frac{1}{4}+\frac{3}{4}}\text{dx}$
$=\int\sqrt{\Big(\text{x}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}\text{dx}$
$=\frac{\big(\text{x}+\frac{1}{2}\big)}{2}\sqrt{\Big(\text{x}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}+\frac{\big(\frac{\sqrt{3}}{2}\big)^2}{2}\times\\\log\Bigg|\Big(\text{x}+\frac{1}{2}\Big)+\sqrt{\Big(\text{x}+\frac{1}{2}\Big)^2+\Big(\frac{\sqrt{3}}{2}\Big)^2}\Bigg|+\text{C}$
$=\Big(\frac{2\text{x}+1}{4}\Big)\sqrt{\text{x}^2+\text{x}+1}+\frac{3}{8}\log\bigg|\Big(\frac{2\text{x}+1}{2}\Big)+\frac{1}{2}\sqrt{\text{x}^2+\text{x}+1}\bigg|+\text{C}$
$\therefore\ \text{I}=\Big(\frac{2\text{x}+1}{4}\Big)\sqrt{\text{x}^2+\text{x}+1}+\frac{3}{8}\log\bigg|2\text{x}+1+\sqrt{\text{x}^2+\text{x}+1}\bigg|+\text{C}$
View full question & answer→Question 1573 Marks
Evalute the following integrals:
$\int\frac{2\cos2\text{x}+\sec^2\text{x}}{\sin2\text{x}+\tan\text{x}-5}\text{dx}$
AnswerLet $\text{I}=\int\frac{2\cos2\text{x}+\sec^2\text{x}}{\sin2\text{x}+\tan\text{x}-5}\text{dx}$
Putting $\sin2\text{x}+\tan\text{x}-5=\text{t}$
$\Rightarrow2\cos2\text{x}+\sec^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(2\cos2\text{x}+\sec^2\text{x})\text{dx}=\text{dt}$
$\therefore\text{I}=\int\frac{1}{\text{t}}\text{dt}$
$=\text{ln}|\text{t}|+\text{C}$
$=\text{ln}|\sin2\text{x}+\tan\text{x}-5|+\text{C} $ $\big[\because\text{t}=\sin2\text{x}+\tan\text{x}-5\big]$
View full question & answer→Question 1583 Marks
Evaluate the following integrals:
$\int\sqrt{9-{\text{x}^2}}\text{dx}$
Answer Let $\text{I}=\int\sqrt{3^2-{\text{x}^2}}$
We know that,
$\int\sqrt{\text{a}^2-\text{x}^2}=\frac{\text{x}}{2}\sqrt{\text{a}^2-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\frac{\text{x}}{\text{a}}+\text{C}$
$\therefore\ \text{I}=\frac{\text{x}}{2}\sqrt{9-\text{x}^2}+\frac{9}{2}\sin^{-1}\frac{\text{x}}{3}+\text{C}$
View full question & answer→Question 1593 Marks
Evaluate the following integrals:
$\int\frac{\text{dx}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
Answer$\int\frac{\text{dx}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
$=\int\frac{\text{dx}}{\text{e}^{\text{x}}+\frac{1}{\text{e}^{\text{x}}}}$
$=\int\frac{\text{e}^{\text{x}}\text{ dx}}{\text{e}^{2\text{x}}+1}$
Let $\text{e}^{\text{x}}=\text{t}$
$\Rightarrow\text{e}^{\text{x}}\text{ dx = dt}$
Now, $\int\frac{\text{e}^{\text{x}}\text{ dx}}{\text{e}^{2\text{x}}+1}$
$=\int\frac{\text{dt}}{1+\text{t}^2}$
$=\tan^{-1}(\text{t})+\text{C}$
$=\tan^{-1}(\text{e}^{\text{x}})+\text{C}$
View full question & answer→Question 1603 Marks
Evaluate the following integrals:
$\int\sqrt{2\text{ax}-\text{x}^2}\text{dx}$
AnswerLet $\text{I}=\int\sqrt{2\text{ax}-\text{x}^2}\text{dx}$
$=\int\sqrt{\text{a}^2+2\text{ax}-\text{x}^2-\text{a}^2}\text{dx}$
$=\int\sqrt{\text{a}^2-(\text{x}^2-2\text{ax}+\text{a}^2)}\text{dx}$
$=\int\sqrt{\text{a}^2-(\text{x}-\text{a})^2}\text{dx}$
$=\Big(\frac{\text{x}-\text{a}}{2}\Big)\sqrt{2\text{ax}-\text{x}^2}+\frac{\text{a}^2}{2}\sin^{-1}\Big(\frac{\text{x}-\text{a}}{\text{a}}\Big)+\text{C}$
View full question & answer→Question 1613 Marks
Evaluate the following integrals:
$\int\frac{1}{\text{x}^{\frac{3}{2}}}\text{dx}$
Answer$\int\frac{1}{\text{x}^\frac{3}{2}}\text{dx}=\int\text{x}\frac{-3}{2}\text{dx}$
$=\int\text{x}^\frac{-3}{2}\text{dx}$
$=\frac{\text{x}^{\frac{-3}{2}+1}}{\frac{-3}{2}+1}+\text{c}$
$=\frac{\text{x}^\frac{-1}{2}}{\frac{-1}{2}}+\text{c}$
$=-2\text{x}\frac{1}{\sqrt{\text{x}}}+\text{c}$
$=\frac{-2}{\sqrt{\text{x}}}+\text{c}$
View full question & answer→Question 1623 Marks
Evaluate the following integrals:$\int\frac{1}{\sqrt{(1-\text{x}^2)\big\{9+\big(\sin^{-1}\text{x}\big)^2\big\}}}\text{ dx}.$
AnswerLet $\text{I}=\int\frac{1}{\sqrt{(1-\text{x}^2)\Big[9+\big(\sin^{-1}\text{x}\big)^2\Big]}}\text{ dx}$
Let $\sin^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{ dx}=\text{dt}$
$\Rightarrow\text{I}=\int\frac{\text{dt}}{\sqrt{(3)^2+\text{t}^2}}$
$\Rightarrow\text{I}=\log\Big|\text{t}+\sqrt{9+\text{t}^2}\Big|+\text{C}$ $\Big[\text{Since }\int\frac{1}{\sqrt{\text{a}^2+\text{x}^2}}\text{ dx}=\log\Big|\text{x}+\sqrt{\text{a}^2+\text{x}^2}\Big|+\text{C}\Big]$
$\text{I}=\log\Big|\sin^{-1}\text{x}+\sqrt{9+\big(\sin^{-1}\text{x}\big)^2}\Big|+\text{C}$
View full question & answer→Question 1633 Marks
If $\int\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}=\text{f(x)}\text{e}^{\text{x}}+\text{C},$ then write the value of f(x).
Answer$\int\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}=\int\Big(\frac{\text{x}}{\text{x}^2}-\frac{1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{dx}$
$=\int\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}$
Consider, $\text{f(x)}=\frac{1}{\text{x}},$ then $\text{f}'(\text{x})=-\frac{1}{\text{x}^2}$
Thus, the given integrand is of the from $\text{e}^{\text{x}}\big|\text{f}(\text{x})+\text{f}'(\text{x})\big|$
Therefore, $\int\Big(\frac{\text{x}-1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{ dx}=\frac{1}{\text{x}}\text{e}^{\text{x}}+\text{C}$
Hence, $\text{f(x)}=\frac{1}{\text{x}}$
View full question & answer→Question 1643 Marks
$\int\frac{2\text{x}-1}{(\text{x}-1)^2}\text{dx}$
Answer$\int\bigg[\frac{2\text{x}-1}{(\text{x}-1)^2}\bigg]\text{dx}$
$=\int\bigg[\frac{2\text{x}-2+2-1}{(\text{x}-1)^2}\bigg]\text{dx}$
$=\int\bigg(\frac{2(\text{x}-1)}{(\text{x}-1)^2}+\frac{1}{(\text{x}-1)^2}\bigg)\text{dx}$
$=2\int\frac{\text{dx}}{\text{x}-1}+\int(\text{x}-1)^{-2}\text{dx}$
$=2\text{ ln}|\text{x}-1|+\frac{(\text{x}-1)^{-2+1}}{-2+1}+\text{C}$
$=2\text{ ln}|\text{x}-1|-\frac{1}{\text{x}-1}+\text{C}$
View full question & answer→Question 1653 Marks
Evaluate the following integrals:$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
Answer$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
Let $\text{t}=\cos^2\text{x}\rightarrow-\text{dt}=2\cos\text{x}\sin\text{x}\text{ dx}$
$\int\frac{\sin2\text{x}}{\sqrt{\cos^4\text{x}-\sin^2\text{x}+2}}\text{ dx}$
$=\int\frac{-1}{\sqrt{\text{t}^2-(1-\text{t})+2}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\text{t}^2+\text{t}+1}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\text{t}^2+\text{t}+\frac{1}{4}+\frac{3}{4}}}\text{ dt}$
$=\int\frac{-1}{\sqrt{\big(\text{t}+\frac{1}{2}\big)^2+\frac{3}{4}}}\text{ dt}$
$=-\log\Big|\Big(\text{t}+\frac{1}{2}\Big)+\sqrt{\text{t}^2+\text{t}+1}\Big|$
$=-\log\Big|\Big(\cos^2\text{x}+\frac{1}{2}\Big)+\sqrt{\cos^4\text{x}+\cos^2\text{x}+1}\Big|+\text{C}$
View full question & answer→Question 1663 Marks
Write a value of $\int\frac{\sin2\text{x}}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin2\text{x}}{\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}}\text{ dx}$
Let $\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}=\text{t}$
$\big(2\text{a}^2\sin\text{x}\cos\text{x}-2\text{b}^2\cos\text{x}\sin\text{x}\big)\text{dx}=\text{dt}$
$2(\text{a}^2-\text{b}^2)\sin\text{x}\cos\text{x dx}=\text{dt}$
$(\text{a}^2-\text{b}^2)\sin2\text{x}\text{ dx}=\text{dt}$
$\text{I}=\frac{1}{\text{a}^2-\text{b}^2}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{\text{a}^2-\text{b}^2}\log\text{t}+\text{C}$
$\text{I}=\frac{1}{\text{a}^2-\text{b}^2}\log\big(\text{a}^2\sin^2\text{x}+\text{b}^2\cos^2\text{x}\big)+\text{C}$
View full question & answer→Question 1673 Marks
Evalute the following integrals:
$\int\frac{1}{\text{x}(3+\log\text{x})}\text{dx}$
AnswerHere, we are considering $\log\text{x}$ as $\log_\text{e}\text{x}$.Let $\text{I}=\int\frac{1}{\text{x}(3+\log\text{x})}\text{dx}$
Putting $\log\tan\text{x}=\text{t}$
$\Rightarrow\frac{1}{\text{x}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{\text{dx}}{\text{x}}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{3+\text{t}}$
$=\log|3+\text{t}|+\text{C}$
$=\log|3+\log\text{x}|+\text{C}\ \big[\because\text{t}=\log\text{x}\big]$
View full question & answer→Question 1683 Marks
Evaluate the following integrals:
$\int\tan\text{x }\sec^4\text{x}\text{dx}$
AnswerLet $\text{I}=\int\tan\text{x }\sec^4\text{x}\text{dx}$ Then
$\text{I}=\int\tan\text{x }\sec^2\text{x}\sec^2\text{x}\text{dx}$
$=\int\tan\text{x}(1+\tan^2\text{x})\sec^2\text{x}\text{dx}$
$\text{I}=\int\big(\tan\text{x}+\tan^3\text{x}\big)\sec^2\text{x}\text{dx}$
Substituting $\tan\text{x}=\text{t}$ and $\sec^2\text{xdx}=\text{dt},$ we get
$\text{I}=\int(\text{t}+\text{t}^3)\text{dt}$
$=\frac{\text{t}^2}{2}+\frac{\text{t}^4}{4}+\text{C}$
$=\frac{\tan^2\text{x}}{2}+\frac{\tan^4}{4}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}\times\tan^2\text{x}+\frac{1}{4}\times\tan^4\text{x}+\text{C}$
View full question & answer→Question 1693 Marks
Evaluate the following integrals:
$\int\frac{1}{(\text{x}+1)(\text{x}^2+2\text{x}+2)}\text{ dx}$
Answer$\int\sqrt{\text{e}^\text{x}-1}\text{ dx}$ Let $\text{I}=\int\frac{1}{(\text{x}+1)(\text{x}^2+2\text{x}+2)}\text{ dx}$ $=\int\frac{1}{(\text{x}+1)\big((\text{x}+1)^2+\text{1}\big)}\text{ dx}$Let $\text{x}+1=\tan\text{u}$
$\Rightarrow\text{dx}=\sec^2\text{u du}$ $\therefore\ \text{I}=\int\frac{\sec^2\text{u}}{\tan\text{u}(\tan^2\text{u}+1)}\text{ du}$ $=\int\frac{\cos\text{u}}{\sin\text{u}}\text{ du}$ $=\log|\sin\text{u}|+\text{C}$ $=\log\Big|\frac{\tan\text{u}}{\sec^2\text{u}}\Big|+\text{C}$ $=\log\bigg|\frac{\text{x}+1}{\sqrt{\text{x}^2+2\text{x}+2}}\bigg|+\text{C}$
View full question & answer→Question 1703 Marks
Evaluate the following integrals:
$\int\cos^5\text{x}\text{ dx}$
Answer$\int\cos^5\text{x}\text{ dx}$
$=\int\cos^4\text{x}\cdot\cos\text{x}\text{ dx}$
$=\int(1-\sin^2\text{x})^2\cos\text{x}\text{ dx}$
Let $\sin\text{x}=\text{t}$
$\cos\text{x}\text{ dx}=\text{dt}$
Now, $\int(1-\sin^2\text{x})^2\cos\text{x}\text{ dx}$
$=\int(1-\text{t}^2)^2\text{ dt}$
$=\int(1+\text{t}^4-2\text{t}^2)\text{dt}$
$=\int\text{dt}+\int\text{t}^4\text{ dt}-2\int\text{t}^2\text{ dt}$
$=\text{t}+\frac{\text{t}^5}{5}-\frac{2\text{t}^3}{3}+\text{C}$
$=\sin\text{x}+\frac{\sin^5\text{x}}{5}-\frac{2}{3}\sin^3\text{x}+\text{C}$
View full question & answer→Question 1713 Marks
Evaluate the following integrals:
$\int\text{x}^3\sin\text{x}^4\text{dx}$
Answer$\int\text{x}^3.\sin\text{x}^4\text{dx}$
Let $\text{x}^4=\text{t}$
$\Rightarrow4\text{x}^3=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{x}^3\text{dx}=\frac{\text{dt}}{4}$
Now, $\int\text{x}^3.\sin\text{x}^4\text{dx}$
$=\frac{1}{4}\int\sin\text{t}\text{ dt}$
$=\frac{1}{4}[-\cos(\text{t})]+\text{C}$
$=\frac{1}{4}\big[-\cos\text{x}^4\big]+\text{C}$
View full question & answer→Question 1723 Marks
Evaluate the following integrals:$\int\text{x}^3\cos\text{x}^2\text{dx}$
AnswerLet $\text{I}=\int\text{x}^3\cos\text{x}^2\text{dx}$
Let $\text{x}^2=\text{t}$
$2\text{x dx = dt}$
$\text{I}=\frac{1}{2}\int\text{t}\cos\text{t dt}$
Using integration by parts,
$=\frac{1}{2}[\text{t}\int\cos\text{t dt}-\int(1\times\int\cos\text{t dt})\text{dt}]$
$=\frac{1}{2}[\text{t}\times\sin\text{t}-\int\sin\text{t dt}]$
$=\frac{1}{2}[\text{t}\sin\text{t}+\cos\text{t}]+\text{C}$
$\text{I}=\frac{1}{2}[\text{x}^2\sin\text{x}^2+\cos\text{x}^2]+\text{C}$
View full question & answer→Question 1733 Marks
Write a value of $\int\frac{\cos\text{x}}{\sin\text{x}\log\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\cos\text{x}}{\sin\text{x}\log\sin\text{x}}\text{ dx}$
$\int\frac{\cot\text{x}}{\log\sin\text{x}}\text{ dx}$
Let $\log\sin\text{x}=\text{t}$
$\cot\text{x dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log\text{t}+\text{C}$
$=\log(\log\sin\text{x})+\text{C}$
View full question & answer→Question 1743 Marks
$\int\frac{1-\cos\text{x}}{1+\cos\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1-\cos\text{x}}{1+\cos\text{x}}\times\text{dx}.$ Then,
$\text{I}=\int\frac{2\sin^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}}\times\text{dx}$
$=\int\frac{\sin^2\frac{\text{x}}{2}}{\cos^2\frac{\text{x}}{2}}\times\text{dx}$
$=\int\tan^2\frac{\text{x}}{2}\text{dx}$
$=\int\Big(\sec^2\frac{\text{x}}{2}-1\Big)\text{dx}$
$=\frac{\tan\frac{\text{x}}{2}}{\frac{1}2{}}-\text{x+c}$
$=2\tan\frac{\text{x}}{2}-\text{x+c}$
View full question & answer→Question 1753 Marks
Write a value of $\int(\text{e}^{\text{x}\log_\text{e}\text{a}}+\text{e}^{\text{a}\log_\text{e}\text{x}})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}\log_\text{e}\text{a}}+\text{e}^{\text{a}\log_\text{e}\text{x}}\text{ dx}$
$=\int(\text{e}\log\text{a}^{\text{x}}+\text{e}\log\text{x}^{\text{a}})\text{dx}$
$=\int(\text{a}^{\text{x}}+\text{x}^{\text{a}})\text{dx}$
$=\int\text{a}^{\text{x}}\text{dx}+\int\text{x}^{\text{a}}\text{dx}$
$=\frac{\text{a}^{\text{x}}}{\log_\text{e}\text{a}}+\frac{\text{x}^{\text{a}+1}}{\text{a}+1}+\text{C}$
$\therefore\ \text{I}=\frac{\text{a}^{\text{x}}}{\log_\text{e}\text{a}}+\frac{\text{x}^{\text{a}+1}}{\text{a}+1}+\text{C}$
View full question & answer→Question 1763 Marks
Evaluate the following intregals:
$\int\frac{\sin2\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\ \text{dx}$
AnswerLet $\text{I}=\int\frac{\sin2\text{x}}{\sin^4\text{x}+\cos^4\text{x}}\ \text{dx}$
Dividing numerator and denominator by $\cos^2\text{x}$
$\text{I}=\int\frac{2\tan\text{x}\sec^2\text{x}}{\tan^4\text{x}+1}\ \text{dx}$
Let $\tan^2\text{x}=\text{t}$
$2\tan\text{x}\sec^2\text{x}\text{ dx}=\text{dt}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2+1}$
$=\tan^{-1}\text{t}+\text{C}$
$\text{I}=\tan^{-1}(\tan^2\text{x})+\text{C}$
View full question & answer→Question 1773 Marks
Evaluate the following integrals:
$\int\frac{1}{\sqrt{1-\text{x}^2}(\sin^{-1}\text{x})^2}\text{dx}$
Answer$\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}(\sin^{-1}\text{x})^2}$
$\text{Let,}\sin^{-1}\text{x}=\text{t}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\frac{1}{\sqrt{1-\text{x}^2}}\text{dx}=\text{dt}$
$\text{Now,}\int\frac{\text{dx}}{\sqrt{1-\text{x}^2}(\sin^{-1}\text{x})^2}$
$=\int\frac{\text{dt}}{\text{t}^2}$
$=\int\text{t}^{-2}\text{dt}$
$=\frac{\text{t}^{-2+1}}{-2+1}+\text{C}$
$=\frac{-1}{\text{t}}+\text{C}$
$=-\frac{1}{\sin^{-1}\text{x}}+\text{C}$
View full question & answer→Question 1783 Marks
Evaluate the following integrals:
$\int\frac{\cos\text{x}}{1+\cos\text{x}}\text{dx}$
Answer$\int\frac{\cos\text{x}}{1+\cos\text{x}}\text{dx}$
$=\int\frac{\cos\text{x}(1-\cos\text{x})}{(1+\cos\text{x})(1-\cos\text{x})}\text{dx}$
$=\int\frac{\cos\text{x}-\cos^2\text{x}}{1-\cos^2\text{x}}\text{dx}$
$=\int\frac{\cos\text{x}-\cos^2\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int\frac{\cos\text{x}}{\sin^2\text{x}}-\frac{\cos^2\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int(\cot\text{x}\text{ cosec x}-\cot^2\text{x})\text{dx}$
$=\int(\cot\text{x}\text{ cosec x}-\text{cosec}^2\text{x}+1)\text{dx}$
$=\int\cot\text{x}\text{ cosec x dx}-\int\text{cosec}^2\text{x dx}+\int1\text{dx}$
$=-\text{cosec x}+\cot\text{x}+\text{x}+\text{C}$
View full question & answer→Question 1793 Marks
Evaluate the following integrals:
$\int\frac{\cos\text{x}}{\sin^2\text{x}+4\sin\text{x}+5}\text{dx}$
Answer$\int\frac{\cos\text{x dx}}{\sin^2\text{x}+4\sin\text{x}+5}$
Let $\sin\text{x = t}$
$\Rightarrow\cos\text{x dx = dt}$
Now, $\int\frac{\cos\text{x dx}}{\sin^2\text{x}+4\sin\text{x}+5}$
$=\int\frac{\text{dt}}{\text{t}^2+4\text{t}+5}$
$=\int\frac{\text{dt}}{\text{t}^2+2\times\text{t}\times2+4+1}$
$=\int\frac{\text{dt}}{(\text{t}+2)^2+1^2}$
$=\frac{1}{1}\tan^{-1}\Big(\frac{\text{t}+2}{1}\Big)+\text{C}$
$=\tan^{-1}(\sin\text{x}+2)+\text{C}$
View full question & answer→Question 1803 Marks
Evaluate the following integrals:
$\int\frac{1-\cos\text{x}}{1+\cos\text{x}}\text{dx}$
Answer$\int\Big(\frac{1-\cos\text{x}}{1+\cos\text{x}}\Big)\text{dx}$
$=\int\frac{(1-\cos\text{x})^2}{1-\cos^2\text{x}}\text{dx}$
$=\int\frac{1+\cos^2\text{x}-2\cos\text{x}}{\sin^2\text{x}}\text{dx}$
$=\int\Big(\frac{1}{\sin^2\text{x}}+\frac{\cos^2\text{x}}{\sin^2\text{x}}-\frac{2\cos\text{x}}{\sin^2\text{x}}\Big)\text{dx}$
$=\int(\text{cosec}^2\text{x}+\cot^2\text{x}-2\cot\text{x}\text{ cosec x})\text{dx}$
$=\int(\text{cosec}^2\text{x}+\text{cosec}^2\text{x}-1-2\cot\text{x cosec x})\text{dx}$
$=\int(2\text{cosec}^2\text{x}-1-2\cot\text{x cosec x})\text{dx}$
$=\int2\text{cosec}^2\text{x dx}-\int1\text{dx}-\int2\cot\text{x cosec x }\text{dx}$
$=-2\cot\text{x}-\text{x}+2\text{cosec x}+\text{C}$
$=2(\text{cosec x}-\cot\text{x})-\text{x + C}$
View full question & answer→Question 1813 Marks
Evaluate the following integrals:
$\int\sqrt{4\text{x}^2-5}\text{dx}$
Answer$\text{I}=\int\sqrt{4\text{x}^2-5}\text{dx}$
$=\int\sqrt{4\Big(\text{x}^2-\frac{5}{4}\Big)}\text{dx}$
$=2\int\sqrt{\text{x}^2-\Big(\frac{\sqrt5}{2}}\Big)^2\text{dx}$
$=2\Big[\frac{\text{x}}{2}\sqrt{\text{x}^2-\frac{5}{4}}-\frac{5}{8}\int\Big|\text{x}+\sqrt{\text{x}^2-\frac{5}{4}}\Big|\Big]+\text{C}$
$\Big[\because\ \int\sqrt{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{x}^2-\text{a}^2}\frac{1}{2}\text{a}^2\int\Big|\text{x}+\sqrt{\text{x}^2-\text{a}^2}\Big|+\text{C}\Big]$
$=\text{x}\sqrt{\text{x}^2-\frac{5}{4}}-\frac{5}{4}\int\bigg|\text{x}+\sqrt{\text{x}^2-\frac{5}{4}}\bigg|+\text{C}$
View full question & answer→Question 1823 Marks
Write the anti-derivative of $\Big(3\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)$
AnswerLet $\text{I}=\int\Big(3\sqrt{\text{x}}+\frac{1}{\sqrt{\text{x}}}\Big)\text{dx}$
$\text{I}=3\sqrt{\text{x}}\text{ dx}+\int\frac{\text{dx}}{\sqrt{\text{x}}}$
$=3\frac{\text{x}^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{\text{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\text{C}$
$=3\frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}+\frac{\text{x}^{\frac{1}{2}}}{\frac{1}{2}}+\text{C}$
$=2\times3\times\frac{\text{x}^\frac{3}{2}}{3}+2\times\frac{\text{x}^{\frac{1}{2}}}{1}+\text{C}$
$=2\Big(\text{x}^{\frac{3}{2}}+\text{x}^{\frac{1}{2}}\Big)+\text{C}$
View full question & answer→Question 1833 Marks
Evaluate the following integrals:
$\int\frac{4\text{x}+3}{\sqrt{2\text{x}^2+3\text{x}+1}}\text{dx}$
Answer$\int\bigg(\frac{4\text{x}+3}{\sqrt{2\text{x}^2+3\text{x}+1}}\bigg)\text{dx}$
$\text{Let }\sqrt{2\text{x}^2+3\text{x}+1}=\text{t}$
$\Rightarrow(4\text{x}+3)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(4\text{x}+3)\text{dx}=\text{dt}$
$\text{Now,}\int\bigg(\frac{4\text{x}+3}{\sqrt{2\text{x}^2+3\text{x}+1}}\bigg)\text{dx}$
$=\int\frac{\text{dt}}{\sqrt{t}}$
$=\int\text{t}^{-\frac{1}{2}}\text{dt}$
$=\Bigg[\frac{\text{t}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\Bigg]+\text{C}$
$=2\sqrt{\text{t}}+\text{C}$
$=2\sqrt{2\text{x}^2+3\text{x}+1}+\text{C}$
View full question & answer→Question 1843 Marks
Evaluate the following integrals:
$\int\sin^3\text{x}\cos^6\text{x}\text{ dx}$
AnswerLet $\text{I}=\int\sin^3\text{x}\cos^6\text{x}\text{ dx}$
Here, power of $\sin\text{x}$ is odd, so we substitute
$\cos\text{x}=\text{t}$
$-\sin\text{x}\text{dt}=\text{dt}$
$\therefore\ \text{I}=\int\sin^2\text{x}\text{ t}^6(-\text{dt})$
$=-\int\big(1-\cos^2\text{x}\big)\text{t}^6\text{dt}$
$=-\int\big(\text{t}^6-\text{t}^8\big)\text{dt}$
$=-\frac{\text{t}^7}{7}+\frac{\text{t}^9}{9}+\text{C}$
$\therefore\ \text{I}=\frac{1}{7}\cos^7\text{x}+\frac{1}{9}\cos^9\text{x}+\text{C}$
View full question & answer→Question 1853 Marks
Evalute the following integrals:
$\int\frac{1}{\cos3\text{x}-\cos\text{x}}\text{dx}$
Answer$\int\frac{1}{\cos3\text{x}-\cos\text{x}}\text{dx}$
$=\int\frac{1}{4\cos^2\text{x}-4\cos\text{x}}\text{dx}\big[\because\cos3\text{x}=\text{4}\cos^3\text{x}\cos\text{x}\big]$
$=\int\frac{1}{4\cos\text{x}(\cos^2\text{x}-1)}\text{dx}$
$=\frac{-1}{4}\int\frac{1}{\cos\text{x}\sin^2\text{x}}\text{dx}$
$=\frac{-1}{4}\int\Big(\frac{\sin^2\text{x}+\cos^2\text{x}}{\cos\text{x}\sin^2\text{x}}\text{dx}\Big)$
$=\frac{-1}{\text{4}}\Big[\int\sec\text{ x dx}+\int\cot\text{x cosec x dx}$
$=\frac{-1}{4}\big(\text{ln }|\sec\text{x}+\tan\text{x}|-\text{cosec x}\big)+\text{C}$
$=\frac{1}{4}\big(\text{cosec x}-\text{ln}|\sec\text{x}+\tan\text{x}|\big)+\text{C}$
View full question & answer→Question 1863 Marks
If $\int\text{e}^{\text{x}}(\tan\text{x}+1)\sec\text{x dx}=\text{e}^{\text{x}}\text{ f}(\text{x})+\text{C},$ then write the value of f(x).
AnswerSince, $\int\text{e}^{\text{x}}\big(\text{f(x)}+\text{f}(\text{x})\big)\text{dx}=\text{e}^{\text{x}}\text{ f}(\text{x})+\text{C}$
We can write the expression as
$\int\text{e}^{\text{x}}(\tan\text{x}+1)\sec\text{x dx}=\text{e}^{\text{x}}\text{ f}(\text{x})+\text{C}$
$\int\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x}\sec\text{x})\text{dx}=\text{f}(\text{x})\text{e}^{\text{x}}+\text{C}$
Comparing with equation $\int\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x}\sec\text{x})\text{dx}=\text{f}(\text{x})\text{e}^{\text{x}}+\text{C}$ with standard equation we have
$\text{f(x)}=\sec\text{x},\text{f}'(\text{x})=\sec\text{x}\tan\text{x}$
Therefore,
$\int\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x}\sec\text{x})\text{dx}=\text{e}^{\text{x}}\sec\text{x}+\text{C}$
Thus, $\text{f(x)}=\sec\text{x}$
View full question & answer→Question 1873 Marks
Evaluate the following integrals:$\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$
$=\int\text{e}^{\text{x}}(\sec\text{x}+\sec\text{x}\tan\text{x})\text{dx}$
Here,$\text{f(x)}=\sec\text{x}$ Put$\text{e}^{\text{x}}\text{f(x)}=\text{t}$
$\Rightarrow\text{f}'(\text{x})=\sec\text{x}\tan\text{x}$
Let $\text{e}^{\text{x}}\sec\text{x}=\text{t}$
Diff. both sides e.r.t.x
$\text{e}^{\text{x}}\sec\text{x}+\text{e}^{\text{x}}\sec\text{x}\tan\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}^{\text{x}}(\sec\text{x}+\tan\text{x})\text{dx = dt}$
$\therefore\int\text{e}^{\text{x}}(\sec\text{x}+\sec\text{x}\tan\text{x})\text{dx}=\int\text{dt}$
$=\text{t}+\text{C}$
$=\text{e}^{\text{x}}\sec\text{x}+\text{C}$
View full question & answer→Question 1883 Marks
Evalute the following integrals:
$\int\frac{\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}}{\text{e}^\text{x}+\text{x}^\text{e}}\text{dx}$
AnswerLet $\text{I}=\int\frac{\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}}{\text{e}^\text{x}+\text{x}^\text{e}}\text{dx}$
Putting $e^x + x^e = t$
$\Rightarrow\text{e}^\text{x}+\text{ex}^{\text{e}-1}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{e}\big(\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}\big)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\big(\text{e}^{\text{x}-1}+\text{x}^{\text{e}-1}\big)\text{dx}=\frac{\text{dt}}{\text{e}}$
$\therefore\text{I}=\frac{1}{\text{e}}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{1}{\text{e}}\text{ ln}|\text{t}|+\text{C}$
$=\frac{1}{\text{e}}\text{ ln}\Big|\text{e}^\text{x}+\text{x}^\text{e}\Big|+\text{C}\big[\because\text{t}=\text{e}^\text{x}+\text{x}^\text{e}\big]$
View full question & answer→Question 1893 Marks
Evaluate the following integrals:
$\int\frac{\sin(2+3\log\text{x})}{\text{x}}\text{ dx}$
Answer$\int\frac{\sin(2+3\log\text{x})}{\text{x}}\text{ dx}$ Let $2+3\log\text{x}=\text{t}$ $\Rightarrow\frac{3}{\text{x}}=\frac{\text{dt}}{\text{dx}}$ $\Rightarrow\frac{\text{dx}}{\text{x}}=\frac{\text{dt}}{3}$ Now, $\int\frac{\sin(2+3\log\text{x})}{\text{x}}\text{ dx}$$=\frac{1}{3}\int\sin\text{t dt}$
$=\frac{1}{3}[-\cos\text{t}]+\text{C}$
$=-\frac{1}{3}\cos(2+3\log\text{x})+\text{C}$
View full question & answer→Question 1903 Marks
Evaluate the following integrals:
$\int\cos\text{x}\sqrt{4-\sin^2\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\cos\text{x}\sqrt{4-\sin^2\text{x}}\text{dx}$
Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x dx}=\text{dt}$
$\Rightarrow\text{I}=\int\sqrt{4-\text{t}^2}\text{dt}$
$=\int\sqrt{2^2-\text{t}^2}\text{dt}$
$=\frac{\text{t}^2}{2}\sqrt{2^2-\text{t}^2}+\frac{4}{2}\sin^{-1}\frac{\text{t}}{2}+\text{C}$
$\therefore\ \text{I}=\frac{1}{2}\sin\text{x}\sqrt{4-\sin^2\text{x}}+2\sin^{-1}\Big(\frac{\sin\text{x}}{2}\Big)+\text{C}$
View full question & answer→Question 1913 Marks
Evaluate the following integrals:
$\int\sqrt{\text{x}-\text{x}^2}\text{dx}$
Answer$\int\sqrt{\text{x}-\text{x}^2}\text{dx}$
$=\int\sqrt{-(\text{x}^2-\text{x})}\text{dx}$
$=\int\sqrt{-\bigg\{\text{x}^2-\text{x}+\Big(\frac{1}{2}\Big)^2-\Big(\frac{1}{2}\Big)^2\bigg\}}\text{dx}$
$=\int\sqrt{\Big(\frac{1}{2}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2}\text{dx}$
$=\Big(\frac{\text{x}-\frac{1}{2}}{2}\Big)\sqrt{\text{x}-\text{x}^2}+\frac{1}{8}\sin^{-1}\bigg(\frac{\text{x}-\frac{1}{2}}{\frac{1}{2}}\bigg)+\text{C}$
$\Big[\because\ \int\sqrt{\text{a}^2-\text{x}^2}\text{dx}=\frac{1}{2}\text{x}\sqrt{\text{a}^2-\text{x}^2}+\frac{1}{2}\text{a}^2\sin^{-1}\frac{\text{x}}{\text{a}}=\text{C}\Big]$
$=\Big(\frac{2\text{x}-1}{4}\Big)\sqrt{\text{x}-\text{x}^2}+\frac{1}{8}\sin^{-1}(2\text{x}-1)+\text{C}$
View full question & answer→Question 1923 Marks
Evaluate the following integrals:
$\int\cot \text{x}. \text{log}\ \sin\text{x dx}$
Answer$\int\cot \text{x}. \text{log}\ \sin\text{x dx}$
$\text{Let}\ \text{log}\sin\text{x} =\text{t}$
$\Rightarrow\frac{1}{\sin\text{x}}\times \cos\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\cot\text{x dx} =\text{dt}$
$\text{Now},\int\cot\text{x}. \log\sin\text{x dx} $
$=\int\text{t}.\text{dt}$
$=\frac{\text{t}^{2}}{2}+\text{C}$
$=\frac{(\text{log}\begin{vmatrix}\sin\text{x}\end{vmatrix})^{2}}{2}+\text{C}$
View full question & answer→Question 1933 Marks
Write a value of $\int\frac{1+\cot\text{x}}{\text{x}+\log\sin\text{x}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{1+\cot\text{x}}{\text{x}+\log\sin\text{x}}\text{ dx}$
Let $\text{x}+\log\sin\text{x}=\text{t}$
$(1+\cot\text{x})\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
Hence,
$\text{I}=\log|\text{x}+\log\sin\text{x}|+\text{C}$
View full question & answer→Question 1943 Marks
Evaluate the following integrals:$\int\frac{\sin\text{x}}{\sqrt{4\cos^2\text{x}-1}}\text{ dx}$
AnswerLet $\text{I}=\int\frac{\sin\text{x}}{\sqrt{4\cos^2\text{x}-1}}\text{ dx}$
Let $2\cos\text{x}=\text{t}$
$\Rightarrow-2\sin\text{x}\text{ dx}=\text{dt}$
$\Rightarrow\sin\text{x}\text{ dx}=-\frac{\text{dt}}{2}$
$\text{I}=-\frac{1}{2}\int\frac{\text{dt}}{\sqrt{\text{t}^2-1}}$
$=-\frac{1}{2}\log\Big|\text{t}+\sqrt{\text{t}^2-1}\Big|+\text{C}$ $\Big[\text{Since }\int\frac{1}{\sqrt{\text{x}^2-\text{a}^2}}\text{ dx}=\log\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|+\text{C}\Big]$
$=-\frac{1}{2}\log\Big|2\cos\text{x}+\sqrt{4\cos^2\text{x}-1}\Big|+\text{C}$
View full question & answer→Question 1953 Marks
Write a value of $\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$
AnswerLet $\text{I}=\int\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}$ $=\int\text{e}^{\text{x}}(\sec\text{x}+\sec\text{x}\tan\text{x})\text{dx}$ Let $\text{e}^{\text{x}}\sec\text{x}=\text{t}$ $(\text{e}^{\text{x}}\sec\text{x}+\text{e}^{\text{x}}\sec\text{x}\tan\text{x})\text{dx}=\text{dt}$ $\text{e}^{\text{x}}\sec\text{x}(1+\tan\text{x})\text{dx}=\text{dt}$ $\therefore\text{I}=\int\text{dt}$ $=\text{t}+\text{C}$$=\text{e}^{\text{x}}\sec\text{x}+\text{C}$ $(\because\text{t}=\text{e}^{\text{x}}\sec\text{x})$
View full question & answer→Question 1963 Marks
Evalute the following integrals:
$\int\frac{1+\tan\text{x}}{1-\tan\text{x}}\text{dx}$
AnswerLet $\text{I}=\int\frac{1+\tan\text{x}}{1-\tan\text{x}}\text{dx}$
$=\int\bigg(\frac{1+\frac{\sin\text{x}}{\cos\text{x}}}{1-\frac{\sin\text{x}}{\cos\text{x}}}\bigg)\text{dx}$
$=\int\Big(\frac{\cos\text{x}+\sin\text{x}}{\cos\text{x}-\sin\text{x}}\Big)\text{dx}$
Putting $\cos\text{x}-\sin\text{x}=\text{t}$
$\Rightarrow(-\sin\text{x}-\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow(\sin\text{x}+\cos\text{x})\text{dx}=-\text{dt}$
$\therefore\text{I}=-\int\frac{1}{\text{t}}\text{dt}$
$=-\text{ln}|\text{t}|+\text{C}$
$=-\text{ln}|\cos\text{x}-\sin\text{x}|+\text{C}\ \big[\because\text{t}\cos\text{x}-\sin\text{x}\big]$
View full question & answer→Question 1973 Marks
Evaluate the following integrals:$\int\text{x}\sin\text{x}\cos\text{x dx}$
AnswerLet $\text{I}=\int\text{x}\sin\text{x}\cos\text{x dx}$
$=\int\frac{\text{x}}{2}(2\sin\text{x} \cos\text{x})\text{dx}$
$=\frac{1}{2}\int\text{x}\sin2\text{x dx}$
Using integration by parts,
$=\frac{1}{2}[\text{x}\int\sin2\text{x dx}-\int(1\times\int\sin2\text{x dx})\text{dx}]$
$=\frac{1}{2}\Big[\text{x}\Big(\frac{-\cos2\text{x}}{2}\Big)-\int\Big(\frac{-\cos2\text{x}}{2}\Big)\text{dx}\Big]$
$=-\frac{1}{4}\text{x}\cos2\text{x}+\frac{1}{4}\int\cos2\text{x dx}$
$\text{I}=-\frac{1}{4}\text{x}\cos2\text{x}+\frac{1}{8}\sin2\text{x}+\text{C}$
View full question & answer→Question 1983 Marks
Evalute the following integrals:
$\int\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{10^\text{x}+\text{x}^{10}}\text{dx}$
AnswerLet $\text{I}=\int\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{10^\text{x}+\text{x}^{10}}\text{dx}\ .....(\text{i})$
Let $10^\text{x}+\text{x}^{10}=\text{t}$ then,
$\text{d}(10^\text{x}+\text{x}^{10})=\text{dt}$
$\Rightarrow(10^\text{x}\log_\text{e}10+10\text{x}^9)\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{(10\text{x}^9+10^\text{x}\log_\text{e}10)}$
Putting $10^x + x^{10} = t$ and $\text{dx}=\frac{\text{dt}}{10\text{x}^9+10^\text{x}\log_\text{e}10}$ in equation (i), we get,
$\text{I}=\frac{10\text{x}^9+10^\text{x}\log_\text{e}10}{\text{t}}\times\frac{\text{dt}}{10\text{x}^9+10^\text{x}\log_\text{e}10}$
$=\int\frac{\text{dt}}{\text{t}}$
$=\log|\text{t}|+\text{C}$
$=\log|10^\text{x}+\text{x}^{10}|+\text{C}$
$\therefore\text{I}=\log|10^\text{x}+\text{x}^{10}|+\text{C}$
View full question & answer→Question 1993 Marks
$\int\frac{1}{1+\cos3\text{x}}\text{dx}$
Answer$\int\frac{\text{dx}}{1+\cos3\text{x}}$
$=\int\frac{(1-\cos3\text{x})}{(1+\cos3\text{x})(1-\cos3\text{x})}\text{dx}$
$=\int\Big(\frac{1-\cos3\text{x}}{1-\cos^23\text{x}}\Big)\text{dx}$
$=\int\Big(\frac{1-\cos3\text{x}}{\sin^23\text{x}}\Big)\text{dx}$
$=\int\text{cosec}^23\text{x}\text{ dx}-\int\text{cosec}3\text{x}\cot3\text{x dx}$
$=-\frac{\cot3\text{x}}{3}+\frac{\text{cosec}3\text{x}}{3}+\text{c}$
$=\frac{1}{3}[\text{cosec}3\text{x}-\cot3\text{x}]+\text{c}$
$=\frac{1}{3}\bigg[\frac{1}{\sin3\text{x}}-\frac{\cos3\text{x}}{\sin3\text{x}}\bigg]+\text{c}$
$=\frac{1}{3}\bigg[\frac{1-\cos3\text{x}}{\sin3\text{x}}\bigg]+\text{c}$
View full question & answer→Question 2003 Marks
$\int(5\text{x}+3)\sqrt{2\text{x}-1}\text{ dx}$
Answer$\text{Let I}=\int(5\text{x}+3)\sqrt{2\text{x}-1}\text{ dx}$
$\text{Putting}\ \ 2\text{x}-1=\text{t}$
$\Rightarrow2\text{x}=\text{t}+1$
$\Rightarrow\text{x}=\frac{\text{t}+1}{2}$
$\&\ 2\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{2}$
$\therefore\text{I}=\int\Big[5\Big(\frac{\text{t}+1}{2}\Big)+3\Big]\times\sqrt{\text{t}}\times\frac{\text{dt}}{2}$
$=\int\Big(\frac{5\text{t}}{5}+\frac{5}{2}+3\Big)\times\frac{\sqrt{\text{t}}\text{ dt}}{2}$
$=\frac{1}{4}\int(5\text{t}+11)\text{t}^\frac{1}{2}\text{ dt}$
$=\frac{1}{4}\int(5\text{t}^\frac{3}{2}+11\text{t}^\frac{1}{2})\text{ dt}$
View full question & answer→