∫[(sec^(3 / 2) θ-sec^(1 / 2) θ) /(2+tan^2 θ)] tanθd θ= +c (a) (1 / √(2)) log[[secθ-√((2 secθ)+1)

∫[(sec^(3 / 2) θ-sec^(1 / 2) θ) /(2+tan^2 θ)] tanθd θ= +c...

$\int\left[\left(\sec ^{(3 / 2)} \theta-\sec ^{(1 / 2)} \theta\right) /\left(2+\tan ^{2} \theta\right)\right] \tan \theta d \theta=$ $+c$ (a) $(1 / \sqrt{2}) \log [[\sec \theta-\sqrt{(2 \sec \theta)+1}] /[\sec \theta+\sqrt{(2 \sec \theta)}+1] \mid$ (b) $(1 / \sqrt{2}) \log [[\sec \theta+\sqrt{(} 2 \sec \theta)+1] /[\sec \theta-\sqrt{(} 2 \sec \theta)+1]$ (c) $(1 / \sqrt{2}) \log \mid[\sec \theta-\sqrt{(} 2 \sec \theta)-1] /[\sec \theta+\sqrt{(2 \sec \theta)-1]} \mid$ (d) $(1 / \sqrt{2}) \log [[\sec \theta+\sqrt{(2 \sec \theta)-1] /[\sec \theta-\sqrt{(2 \sec \theta)}-1]}]$

$\int\left[\left(\sec ^{(3 / 2)} \theta-\sec ^{(1 / 2)} \theta\right) /\left(2+\tan ^{2} \theta\right)\right] \tan \theta d \theta=$
$+c$
(a) $(1 / \sqrt{2}) \log [[\sec \theta-\sqrt{(2 \sec \theta)+1}] /[\sec \theta+\sqrt{(2 \sec \theta)}+1] \mid$
(b) $(1 / \sqrt{2}) \log [[\sec \theta+\sqrt{(} 2 \sec \theta)+1] /[\sec \theta-\sqrt{(} 2 \sec \theta)+1]$
(c) $(1 / \sqrt{2}) \log \mid[\sec \theta-\sqrt{(} 2 \sec \theta)-1] /[\sec \theta+\sqrt{(2 \sec \theta)-1]} \mid$
(d) $(1 / \sqrt{2}) \log [[\sec \theta+\sqrt{(2 \sec \theta)-1] /[\sec \theta-\sqrt{(2 \sec \theta)}-1]}]$

Key Concepts

Integration of Trigonometric Functions

Integrating trigonometric functions, especially when they involve fractional exponents or are combined in a nonstandard form, often requires a blend of algebraic manipulation and the use of trigonometric identities. Understanding how to decompose or rearrange such integrals is key to solving them effectively.

Trigonometric Identities

Trigonometric identities, such as the relationship between the secant and tangent functions (e.g., tan²? + 1 = sec²?), are fundamental in rewriting the integrand into simpler forms. These identities allow for the transformation and simplification of expressions, which is essential in setting up an integration problem for further manipulation.

U-Substitution

U-substitution is a core technique in integration where a part of the integrand is substituted with a new variable to simplify the differential form. This method is particularly useful when the derivative of the chosen substitution appears in the integrand, thereby transforming a complex expression into one that is readily integrable.

Logarithmic Integration

Logarithmic integration arises when the integrand's structure, such as a derivative divided by a function, leads to an antiderivative in the form of a natural logarithm. Recognizing these forms and knowing how to manipulate them are essential skills in handling integrals that result in logarithmic expressions.

Transcript

00:01 I've been able to find out this increase here we have, so let's go with that.

00:06 So this put your substitution, we just go with that.

00:09 So we use here 10 square theta, that equals x square theta, negative 1.

00:17 So coming after we have x theta, with a part 3 over 2, negative root, set theta over set theta, then text square theta, positive 1, so tect theta 10 theta d theta now a substitution here we have so for substitution we'll use here u equals sec theta for d u over d theta that is tech theta 10 theta we get from here d theta equals 1 over tech theta 10 theta d u.

01:02 Now the matter to be here we have d2 we have negative root u over we have u square plus 1 d now we substitute here uh v equals root u so next position is coming up to be v equals root u so d u so du equals 2 root u d b so 1 over root u equals 1 over b so u s s s s s s sq equal equal of 4 will become here integral i'm reading in blue color that is given as 2 v2 negative 1 over 2d2 of 4 plus 1 d equal here next i just call this to factorize the denominator so when you factorize the denominator and so let's first solve this here we have we have or solve this part here first i solve this in red so it's coming out to be integral v square negative 1 over v square negative root 2v positive 1 times we have v squared positive root 2v positive positive positive plus we have and then dm we go to this now so let's now solve this we get it as 2v negative root 2 over we have 2 to the 5 3 over 2 v2 square negative root 2 we perform here partial question decomposition plus 1 negative 2v plus root 2 over 2 to part 3 over 2 and then we have v square plus root 2 v plus 1 this coming out to be d b next is all called this and simplify we get as 1 over 2 to of the part 3 over 2v negative root 2 over v square negative root 2 v plus 1 db negative 2 2 2b negative 1 over 2 2 b 2b plus root 2 over we have v square plus root 2b plus 1 db so this we have got here now let's simplify this so we get now we just to simplify this integral here the substitution you can simplify this so by this we'll get it as there'll be ln v square negative root to v plus 1 and when you simplify this one here substitution you will get ln v square plus root to v plus plus one now next we just plug in the solve integral so you would get ln v square negative root to be plus 1 over 2 to the part p over 2 negative negative ln you get v squared positive root 2v positive 1 over 2 2 2 2 2 2 2 2 2 2 2 plug in the solid integral you will have so you will get 2 times integral v square negative 1 over b2 plus 1 d b2 that is given as ln v -square negative root 2 v plus 1 over root 2 negative ln v square positive root 2 plus 1 over root 2.

05:23 So, we have now...

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