Mathematics · 2019
JEE Main · 9 April 2019, Shift 2 · Q74
If ∫ e^sec x( sec x tan x f(x)+( sec x tan x+ sec^2 x)) d x = e^sec x f(x)+ C, then a possible choice of; f(x) is:
If
$$\begin{aligned}
& \int \mathrm{e}^{\sec x}\left(\sec x \tan x f(x)+\left(\sec x \tan x+\sec ^2 x\right)\right) \mathrm{d} x \\
& =\mathrm{e}^{\sec x} f(x)+\mathrm{C}, \text { then a possible choice of } \\
& f(x) \text { is : }
\end{aligned}
$$
Official answer
(2)
$\displaystyle \sec x+\tan x+\frac{1}{2}$
More from Indefinite Integration
- If ∫ 1/x √((1-x)/(1+x)) d x=g(x)+c, g(1)=0, then g(1/2) is equal to:2022
- If ∫ (cos x d x)/(sin^3 x(1+ sin^6 x)^2 / 3)=f(x)(1+ sin^6 x)^1 / λ+c where c is a constant of integration, then λ f((π)/3) is equal to:2020
- Let f(x)=∫ (d x)/((3+4 x^2) √(4-3 x^2)),|x|<2/(√3). If f(0)=0 and f(1)=1/(α β) tan^-1((α)/(β)), α, β>0, then α^2+β^2 is equal to ____.2023
- If ∫ √(sec 2 x-1) d x=α log_e| cos 2 x+β+√(cos 2 x(1+ cos 1/(β) x))|+ constant, then β-α is equal to ____.2023
- The integral ∫(x/(x sin x+ cos x))^2 d x is equal to (where C is a constant of integration):2020
- Let I(x)=∫ (x^2(x sec^2 x+ tan x))/((x tan x+1)^2) d x. If I(0)=0, then I((π)/4) is equal to2023
- The integral ∫ ((x^8-x^2) d x)/((x^12+3 x^6+1) tan^-1(x^3+1/x^3)) is equal to:2024
- If ∫ sin^-1(√(x/(1+x))) d x=A(x) tan^-1(√x)+ B(x)+C, where C is a constant of integration, then the ordered pair (A(x), B(x)) can be:2020
JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.